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ELEMENTS 


OF 


GEOMETRY  AND  TRIGONOMETRY. 


TRANSLATED  FROM  THE  FRENCH  OF 


A.  M.  LEGENDRE, 


MEMBER  OF  THE  INSTITUTE  AND  OF  THE  LEGION  OF  HONOUR,  AND  OF  THE  ROYAL 
SOCIETIES  OF  LONDON  AND  EDINBURGH,  &C. 


BY  DAVID  BREWSTER,  LL.  D. 

FKLLOW   OF   THE    ROYAL    SOCIETY    OF    LONDON,   AND   SECRETARY   TO  THE    ROYAL  SOCIETY   Of 
EDINBURGH,  &C.  &C. 


REVISED  AND  ABRIDGED 

BY  CHARLES  DAVIES,  :| 

PROFESSOR  OF  MATHEMATICS  * 

IN  THE 

MILITARY  .ACADEMY, 

AND 

AUTHOR  OF  tHE  COMMON  SCHOOL  ARITHMETIC,  DESCRIPTIVE  GEOMETRY, 

ELEMENTS  OF  SURVEYING,  AND  A  TREATISE  ON 

SHADOWS  AND  PERSPECTIVE. 


FOURTH    EDITION. 


NEW-YORK: 

PUBLISHED  BY  HARPER  AND  BROTHERS, 

NO.  82  CLIFF-STREET. 


Stereotyped  by  A.  Chandler. 

1834. 


.v>....^v,^,.0     ^■^"'^^^ 


Entered  according  to  the  Act  of  Congress,  in  the  year  one  thousand  eight 
hundred  and  thirty-four,  by  Charles  Da  vies,  in  the  Clerk's  Office  of  the  Dis- 
trict Court  of  the  United  States,  for  the  Southern  District  of  New- York. 


PREFACE 


TO  THE  FOURTH  AMERICAN  EDITION. 

The  Editor,  in  offering  to  the  public  Dr.  Brewster's 
translation  of  Legendre's  Geometry  under  its  present 
form,  is  fully  impressed  with  the  responsibility  he 
assumes  in  making  alterations  in  a  work  of  such  de- 
served celebrity. 

In  the  original  work,  as  well  as  in  the  translations 
of  Dr.  Brewster  and  Professor  Farrar,  the  proposi- 
tions are  not  enunciated  in  general  terms,  but  with 
reference  to,  and  by  the  aid  of,  the  particular  diagrams 
used  for  the  demonstrations.  It  is  believed  that  this 
departure  from  the  method  of  Euclid  has  been  gene- 
rally regretted.  The  propositions  of  Geometry  are 
general  truths,  and  as  such,  should  be  stated  in  gene- 
ral terms,  and  without  reference  to  particular  figures. 
The  method  of  enunciating  them  by  the  aid  of  particu- 
lar diagrams  seems  to  have  been  adopted  to  avoid  the 
difiiculty  which  beginners  experience  in  comprehend- 
ing abstract  propositions.  But  in  avoiding  this  diffi- 
culty, and  thus  lessening,  at  first,  the  intellectual 
labour,  the  faculty  of  abstraction,  which  it  is  one  of 
the  primary  objects  of  the  study  of  Geometry  to 
strengthen,  remains,  to  a  certain  extent,  unimproved. 

3037  lis 


iv  :*  PREFACE. 

-Vu^  Besides  the  alterations  in  the  enunciation  of  the 
propositions,  others  of  considerable  importance  have 
also  been  made  in  the  present  edition.  The  propo- 
sition in  Book  V.,  which  proves  that  a  polygon  and 
circle  may  be  made  to  coincide  so  nearly,  as  to  differ 
from  each  other  by  less  than  any  assignable  quantity, 
has  been  taken  from  the  Edinburgh  Encyclopedia. 
It  is  proved  in  the  corollaries  that  a  polygon  of  an 
infinite  number  of  sides  becomes  a  circle,  and  this 
principle  is  made  the  basis  of  several  important  de- 
monstrations in  Book  VIII. 

Book  II.,on  Ratios  and  Proportions,  has  been  partly 
adopted  from  the  Encyclopedia  Metropolitana,  and 
will,  it  is  believed,  supply  a  deficiency  in  the  original 
work. 

Very  considerable  alterations  have  also  been  made 
in  the  manner  of  treating  the  subjects  of  Plane  and 
Spherical  Trigonometry.  It  has  also  been  thought 
best  to  pubhsh  with  the  present  edition  a  table  of 
logarithms  and  logarithmic  sines. 

Military  Academy, 

West  Point,  March,  1834. 


CONTENTS. 

BOOK  I. 

The  principles, 9 

*  ^^     ^  BOOK  II. 

Ratios  and  Proportions, 34 

BOOK  III. 
The  Circle  and  the  Measurement  of  Angles,        -         -         -         41 
Problems  relating  to  the  First  and  Third  Books,  -         -         57 

BOOK  IV. 
The  Proportions  of  Figures  and  the  Measurement  of  Areas,  -         68 
Problems  relating  to  the  Fourth  Book,        -         -         .         -         98 

BOOK  V. 
Regular  Polygons  eind  the  Measurement  of  the  Circle,  -       109 

BOOK  VI. 
Planes  and  SoHd  Angles, 126 

BOOK  VII. 
Polyedrons, 142 

BOOK  VIII. 
The  three  round  bodies, 166 

BOOK  IX. 
Of  Spherical  Triangles  and  Spherical  Polygons,  -         -        -       186 

APPENDIX. 
The  regular  Polyedrons,  -        -        ...        -      205 


VI 


CONTENTS. 


Tan 


PLANE  TRIGONOMETRY. 
Division  of  the  Circumference, 
General  Ideas  relating  to  the  Trigonometrical  Lines, 
Theorems  and  Formulas  relating  to  the  Sines,  Cosines, 
gents,  (fee.         ...... 

Construction  and  Description  of  the  Tables, 
Description  of  Table  of  Logarithms,   - 
Description  of  Table  of  Logarithmic  Sines, 
Principles  for  the  Solution  of  Rectilineal  Triangles, 
Solution  of  Rectilineal  Triangles  by  Logarithms, 
Solution  of  Right  angled  Triangles, 
Solution  of  Triangles  in  general,         .         -         - 

SPHERICAL  TRIGONOMETRY. 
First  principles,       _-.----- 

Napier's  Circular  Parts, 

Solution  of  Right  angled  Spherical  Triangles  by  Logarithms; 
Quadrantal  Triangles,  i  -  -  -  .  -  . 
Solution  of  Oblique  angled  Triangles  by  Logarithms,  - 


207 
208 

215 
223 
224 
228 
231 
235 
237 
238 

246 
252 
255 
257 
259 


AN   INDEX 


SHOWING  THE  PROPOSITIONS    OF   LEGENDRE    WHICH    CORRESPOND  TO 
THE  PRINCIPAL  PROPOSITIONS  OF  THE  FIRST  SIX  BOOKS  OF  EUCLID. 


Euclid. 

Legendre. 

Euclid. 

Legendre. 

Euclid. 

Legendre. 

Book  I. 

Book  I. 

Cor.2.  of  32 
33 
34 

35 

Prop.       27 
30 

28 

Prop.       26 

28 
29 

31 

Prop.           15 
5 
5 

5  Cor.2.  ?To 
i    &3.    r^ 

Prop.         4 
5 

Cor.  of      5 
6 

Prop.         5 
11 

Cor.  of    11 
12 

Book  IV. 

1 

8 
13 

10 
1 

36 
37 

1 
Cor  2  of  2 

Book  IV. 

14 

3 

38 

Cor.  2.  of  2 

35 

28 

15 

Cor.  l.<  ,. 

&2.   \^^ 

16 

17 

4 

4 
47 

2 
11 

8 

36 

30 

Sch.  of     4 

5  Cor.  of 25 
I             25 

Book  VI. 

5  Cor.  1.  of  4 
I  Cor.  of     6 

Book  II. 

1 

4 

18 

13 

12 

13 

19 
20 
21 
24 

13 

13 

12 

2 

S                15 

i                 16 

17 

18 

7 
8 
9 

Book  III. 

Book  III. 

3 

4 

Prop.         3 

Prop.         6 

25 

9 

10 

Cor.  of      7 

5 

19 

26 

6 

11 

Cor.  of    14 

6 

20 

27 

Cor'.  1.  of  19 

12 

Cor.  of    14 

8 

22 

28 

Cor.  2.  of  19 

14 

8 

14 

)                25 
I  Cor.  of   15 

29 

\  Cor.2.  & 
I    4.  of  20 

15 

2 

15 

18 

9 

19 

25 

30 

22 

20 

18 

20 

5                 26 

I                27' 

ICor.l.of32 

26 

21 

Cor.  of    18 

ELEMENTS  OF  GEOMETRY. 


BOOK  I. 

THE  PRINCIPLES. 

Definitions, 

1.  Geometry  is  the  science  which  has  for  its  object  the 
measurement  of  extension. 

Extension  has  three  dimensions,  length,  breadth,  and  height, 
or  thickness. 

2.  A  line  is  length  without  breadth,  or  thickness. 

The  extremities  of  a  line  are  called  points :  a  point,  there- 
fore, has  neither  length,  breadth,  nor  thickness,  but  position 
only. 

3.  A  straight  line  is  the  shortest  distance  from  one  point  to 
another. 

4.  Every  line  which  is  not  straight,  or  composed  of  straight 
lines,  is  a  curved  line. 


Thus,  AB  is  a  straight  line ;  ACDB  is  a 
broken  line,  or  one  composed  of  straight  A.^ 
lines ;  and  AEB  is  a  curved  line. 


The  word  line,  when  used  alone,  will  designate  a  straight 
line  ;  and  the  word  curve,  a  curved  line. 

5.  A  surface  is  that  which  has  length  and  breadth,  without 
height  or  thickness. 

6.  A  plane  is  a  surface,  in  which,  if  two  points  be  assumed 
at  pleasure,  and  connected  by  a  straight  line,  that  line  will  lie 
wholly  in  the  surface. 

7.  Every  surface,  which  is  not  a  plane  surface,  or  composed 
of  plane  surfaces,  is  a  curved  surface. 

8.  A  solid  or  body  is  that  which  has  length,  breadth,  and 
thickness  ;  and  therefore  combines  the  three  dimensions  of 
extension. 

2 


10 


GEOMETRY. 


9.  When  two  straight  lines,  AB,  AC,  meet 
each  other,  their  inchnation  or  opening  is  call- 
ed an  angle,  which  is  greater  or  less  as  the 
lines  are  moreor  less  inclined  or  opened.  The 
point  of  intersection  A  is  the  vertex  of  the  j^ 
angle,  and  the  lines  AB,  AC,  are  its  sides. 

The  angle  is  sometimes  designated  simply  by  the  letter  at 
the  vertex  A ;  sometimes  by  the  three  letters  BAC,  or  CAB, 
the  letter  at  the  vertex  being  always  placed  in  the  middle. 

Angles,  like  all  other  quantities,  are  susceptible  of  addition, 
subtraction,  multiplication,  and  division. 


Thus  the  angle  DCE  is  the  sum  of 
the  two  angles  DCB,  BCE ;  and  the  an- 
gle DCB  is  the  difference  of  the  two 
angles  DCE,  BCE. 


10.  When  a  straight  line  AB  meets  another 
straight  line  CD,  so  as  to  make  the  adjacent 
angles  BAC,  BAD,  equal  to  each  other,  each 
of  those  angles  is  called  a  right  angle  ;  and  the 
line  AB  is  said  to  be  perpendicular  to  CD.       : 


11.  Every  angle  BAC,  less  than  dP^ 
right  angle,  is  an  acute  angle ;   and 
every  angle  DEF,  greater  than  a  right 
angle,  is  an  obtuse  angle. 


-F 


12.  Two  lines  are  said  to  be  parallel,  when 

being  situated  in  the  same  plane,  they  cannot 
meet,  how  far  soever,  either  way,  both  of  them 
be  produced. 

13.  A  plane  figure  is  a  plane  terminated  on 
all  sides  by  lines. 

If  the  lines  are  straight,  the  space  they  enclose 
is  called  a  rectilineal  figure,  or  polygon,  and  the 
lines  themselves,  taken  together,  form  the  contour, 
or  perimeter  of  the  polygon. 

14.  The  polygon  of  three  sides,  the  simplest  of  all,  is  called 
a  triangle  ;  that  of  four  sides,  a  quadrilateral ;  that  of  five,  a 
pentagon ;  that  of  six,  a  hexagon  ;  that  of  seven,  a  heptagon ; 
that  of  eight,  an  octagon  ;  that  of  nine^  a  nonagon ;  that  of  ten,  a 
decagon ;  and  that  of  twelve,  a  dodecagon. 


BOOK  I. 


11 


15.  An  equilateral  triangle  is  one  which  has  its  three  sides 
equal ;  an  isosceles  triangle,  one  which  has  two  of  its  sides 
equal ;  a  scalene  triangle,  one  which  has  its  three  sides  unequal. 

16.  A  right-angled  triangle  is  one  which 
has  a  right  angle.  The  side  opposite  the 
right  angle  is  called  the  hypothenuse.  Thus, 

^in  the  triangle  ABC,  right-angled  at  A,  the 
side  BC  is  the  hypothenuse. 

1 7.  Among  the  quadrilaterals,  we  distinguish  : 

The  square,  which  has  its  sides  equal,  and  its  an- 
gles right-angles. 


The  rectangle,  which  has  its  angles  right  an- 
gles, without  having  its  sides  equal. 

The  parallelogram,  or  rhomboid,  which 
has  its  opposite  sides  parallel. 


The  rhombus,  or  lozenge,  which  has  its  sides  equal, 
without  having  its  angles  right  angles. 


And  lastly,  the  trapezoid,  only  two  of  whose  sides 
are  parallel. 


18.  A  diagonal  is  a  line  which  joins  the  ver- 
tices of  two  angles  not  adjacent  to  each  other. 
Thus,  AF,  AE,  AD,  AC,  are  diagonals. 


19.  Axi  equilateral  polygon  is  one  which  has  all  its  sides 
equal ;  an  equiangular  polygon,  one  which  has  all  its  angles 
equal. 

20.  Two  polygons  are  mutually  equilateral,  when  they  have 
their  sides  equal  each  to  each,  and  placed  in  the  same  order ; 


12  R  f r      GEOMETRY. 

that  is  to  say,  when  following  their  perimeters  in  the  same  di- 
rection, the  first  side  of  the  one  is  equal  to  the  first  side  of  the 
other,  the  second  of  the  one  to  the  second  of  the  other,  the 
third  to  the  third,  and  so  on.  The  phrase,  mutually  equian- 
gular, has  a  corresponding  signification,  with  respect  to  the 
angles. 

In  both  cases,  the  equal  sides,  or  the  equal  angles,  are  named 
homologous  sides  or  angles. 


Definitions  of  terms  employed  in  Geometry, 

An  axiom  is  a  self-evident  proposition. 

A  theorem  is  a  truth,  which  becomes  evident  by  means  of  a 
train  of  reasoning  called  a  demonstration, 

A  problem  is  a  question  proposed,  which  requires  a  solu- 
tion, 

A  lemma  is  a  subsidiary  truth,  employed  for  the  demonstra- 
tion of  a  theorem,  or  the  solution  of  a  problem. 

The  common  name,  proposition,  is  applied  indifferently,  to 
theorems,  problems,  and  lemmas. 

A  corollary  is  an  obvious  consequence,  deduced  from  one  or 
several  propositions. 

A  scholium  is  a  remark  on  one  or  several  preceding  propo- 
sitions, which  tends  to  point  out  their  connexion,  their  use,  their 
restriction,  or  their  extension. 

A  hypothesis  is  a  supposition,  made  either  in  the  enunciation 
of  a  proposition,  or  in  the  course  of  a  demonstration. 


Explanation  of  the  symbols  to  be  employed. 

The  sign  =  is  the  sign  of  equality;  thus,  the  expression 
A=B,  signifies  that  A  is  equal  to  B. 

To  signify  that  A  is  smaller  than  B,  the  expression  A<-B 
is  used. 

To  signify  that  A  is  greater  than  B,  the  expression  A>B 
is  used  ;  the  smaller  quantity  being  always  at  the  vertex  of  the 
angle. 

The  sign   +  is  called  plus :  it  indicates  addition. 

The  sign  —  is  called  minus  :  it  indicates  subtraction. 
Thus,  A  +  B,  represents  the  sum  of  the  quantities  A  and  B ; 
A — B  represents  their  difference,  or  what  remains  after  B  is 
taken  from  A  ;  and  A — B  +  C,  or  A  +  C — B,  signifies  that  A 
and  C  are  to  be  added  together,  and  that  B  is  to  be  subtracted 
from  their  sum. 


BOOK  I.  13 

The  sign  x  indicates  multiplication  :  thus,  A  x  B  represents 
the  product  of  A  and  B.  Instead  of  the  sign  x ,  a  point  is 
sometimes  employed  ;  thus,  A.B  is  the  same  thing  as  A  x  B. 
The  same  product  is  also  designated  without  any  intermediate 
sign,  by  AB  ;  but  this  expression  should  not  be  employed,  when 
there  is  any  danger  of  confounding  it  with  that  of  the  line  AB, 
which  expresses  the  distance  between  the  points  A  and  B. 

The  expression  A  x  (B  +  C — D)   represents  the  product  of 
A  by  the  quantity  B  +  C — D.     If  A  +  B  were  to  be  multiplied 
by  A — B  +  C,  the  product  would  be   indicated  thus,  (A  +  B)x 
(A — B  +  C),  whatever  is  enclosed  within  the  curved  lines,  being 
considered  as  a  single  quantity. 

A  number  placed  before  a  line,  or  a  quantity,  serves  as  a 
multiplier  to  that  line  or  quantity ;  thus,  3AB  signifies  that 
the  line  AB  is  taken  three  times ;  |  A  signifies  the  half  of  the 
angle  A. 

The  square  of  the  line  AB  is  designated  by  AB^ ;  its  cube 
by  ABl  What  is  meant  by  the  square  and  cube  of  a  line,  will 
be  explained  in  its  proper  place. 

The  sign  V  indicates  a  root  to  be  extracted  ;  thus  \/2 
means  the  square-root  of  2  ;  V  A  x  B  means  the  square-root  of 
the  product  of  A  and  B. 

Axioms. 

1.  Things  which  are  equal  to  the  same  thing,  are  equal  to 
each  other. 

2.  If  equals  be  added  to  equals,  the  wholes  will  be  equal. 

3.  If  equals  be  taken  from  equals,  the  remainders  will  be 
equal. 

4.  If  equals  be  added  to  unequals,  the  wholes  will  be  un- 
equal. 

5.  If  equals  be  taken  from  unequals,  the  remainders  will  be 
unequal. 

6.  Things  which  are  double  of  the  same  thing,  are  equal  to 
each  other. 

7.  Things  which  are  halves  of  the  same  thing,  are  equal  to 
each  other. 

8.  The  whole  is  greater  than  any  of  its  parts. 

9.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

10.  All  right  angles  are  equal  to  each  other. 

11.  From  one  point  to  another  only  one  straight  line  can  be 
drawn. 

12.  Through  the  same  point,  only  one  straight  line  can  be 
drawn  which  shall  be  parallel  to  a  given  line. 

13.  Magnitudes,  which  being  applied  to  each  other,  coincide 
throughout  their  whole  extent,  are  equal. 

B 


14 


GEOMETRY. 


PROPOSITION   I.    THEOREM. 

If  one  straight  line  meet  another  straight  line,  the  sum  of  the 
two  adjacent  angles  will  he  equal  to  two  right  angles. 

Let  the  straight  line  DC  meet  the  straight 
line  AB  at  C,  then  will  the  angle  ACD  + 
the  angle  DCB,  be  equal  to  two  right  angles. 

At  the  point  C,  erect  CE  perpendicular  to 
AB.  The  angle  ACD  is  the  sum  of  the  an-•^^ 
gles  ACE,  ECD:  therefore  ACD -f  DCB  is 
the  sum  of  the  three  angles  ACE,  ECD,  DCB :  but  the  first 
of  these  three  angles  is  a  right  angle,  and  the  other  two 
make  up  the  right  angle  ECB  ;  hence,  the  sum  of  the  two  an- 
gles ACD  and  DCB,  is  equal  to  two  right  angles. 

Cor.  1.     If  one  of  the  angles  ACD,  DCB,  is  a  right  angle, 
the  other  must  be  a  right  angle  also. 

Cor,  2.  If  the  line  DE  is  perpendicular 
to  AB,  reciprocally,  AB  will  be  perpendicu- 
lar to  DE. 

For,  since  DE  is  perpendicular  to  AB,  the 
angle  ACD  must  be  equal  to  its  adjacent  an- 
gle DCB,  and  both  of  them  must  be  right 
angles  (Def  10.).  But  since  ACD  is  a 
right  angle,  its  adjacent  angle  ACE  must  also  be  a  right  angle 
(Cor.  1.).  Hence  the  angle  ACD  is  equal  to  the  angle  ACE, 
(Ax.  10.)  :  therefore  AB  is  perpendicular  to  DE. 

Cor,  3.  The  sum  of  all  the  successive 
angles,  BAC,  CAD,  DAE,  EAF,  formed 
on  the  same  side  of  the  straight  line  BF, 
is  equal  to  two  right  angles  ;  for  their  sum 
is  equal  to  that  of  the  two  adjacent  an- 
gles, BAC,  CAR  ^ 


PROPOSITION  II.    THEOREM. 

Tv)o  straight  lines,  which  have  two  points  common,  coincide  with 
#  •."    eac^  other  throughout  their  whole  extent,  and  form  one  and 
.*r-.  the  same  straight  line. 

Let  A  and  B  be  the  two  common 
points.  In  the  first  place  it  is  evident 
that  the  two  lines  must  coincide  entirely 
between  A  and  B,  for  otherwise  there 
would  be  two  straight  lines  between  A 
and  B,  which  is  impossible  (Ax.  11.).  Sup- 


JL     B 


BOOK  I.  15 

pose,  however,  that  on  being  produced,  these  lines  begin  to 
separate  at  C,  the  one  becoming  CD,  the  other  CE.  From 
the  point  C  draw  the  Hne  CF,  making  with  AC  the  right  angle 
ACF.  Now,  since  ACD  is  a  straight  line,  the  angle  FCD  will 
be  a  right  angle  (Prop.  I.  Cor.  1.) ;  and  since  ACE  is  a  straight 
line,  the  angle  FCE  will  likewise  be  a  right  angle.  Hence,  the 
angle  FCD  is  equal  to  the  angle  FCE  (Ax.  10.);  which  can 
only  be  the  case  when  the  lines  CD  and  CE  coincide :  there- 
fore, the  straight  lines  which  have  two  points  A  and  B  com- 
mon, cannot  separate  at  any  point,  when  produced ;  hence  they . 
form  one  and  the  same  straight  line. 

PROPOSITION  III.    THEOREM. 

If  a  straight  line  meet  two  other  straight  lines  at  a  common 
point,  making  the  sum  of  the  two  adjacent  angles  equal  to  two 
right  angles,  the  two  straight  lines  which  are  met,  will  form 
one  and  the  same  straight  line. 

Let  the  straight  line  CD  meet  the 
two  lines  AC,  CB,  at  their  common 
point  C,  making  the  sum  of  the  two 
adjacent  angles  DCA,  DCB,  equal  to  57 
two  right  angles ;  then  will  CB  be  the 
prolongation  of  AC,  or  AC  and  CB 
will  form  one  and  the  same  straight  line. 

For,  if  CB  is  not  the  prolongation  of  AC,  let  CE  be  that  pro- 
longation: then  the  line  ACE  being  straight,  the  sum  of  the 
angles  ACD,  DCE,  will  be  equal  to  two  right  angles  (Prop.  I.). 
But  by  hypothesis,  the  sum  of  the  angles  ACD,  DCB,  is  also 
equal  to  two  right  angles:  therefore,  ACD  +  DCE  must  be  equal 
to  ACD  +  DCB ;  and  taking  away  the  angle  ACD  from  each, 
there  remains  the  angle  DCE  equal  to  the  angle  DCB,  which 
can  only  be  the  case  when  the  lines  CE  and  CB  coincide  ; 
hence,  AC,  CB,  form  one  and  the  same  straight  line. 

PROPOSITION  IV.    THEOREM. 

When  two  straight  lines  intersect  each  other,  the  opposite  or  ver- 
tical  angles,  which  they  form,  are  equal. 


16 


GEOMETRY. 


Let  AB  and  DE  be  two  straight^ 
lines,  intersecting  each  other  at    C  ; 
then  will  the  angle  ECB   be  equal  to 
the  angle  ACD,  and  the  angle  ACE  to 
the  angle  DCB.  TO  B" 

For,  since  the  straight  line  DE  is  met  by  the  straight  line 
AC,  the  sum  of  the  angles  ACE,  ACD,  is  equal  to  two  right 
angles  (Prop.  L) ;  and  since  the  straight  line  AB,  is  met  by  the 
straight  line  EC,  the  sum  of  the  angles  ACE  and  ECB,  is  equal 
to  two  right  angles:  hence  the  sum  ACE  +  ACD  is  equal  to 
the  sum  ACE  +  ECB  (Ax.  1.).  Take  away  from  both,  the  com- 
mon angle  ACE,  there  remains  the  angle  ACD,  equal  to  its 
opposite  or  vertical  angle  ECB  (Ax.  3.). 

Scholium,  The  four  angles  formed  about  a  point  by  two 
straight  lines,  which  intersect  each  other,  are  together  equal  to 
four  right  angles  :  for  the  sum  of  the  two  angles  ACE,  ECB, 
is  equal  to  two  right  angles  ;  and  the  sum  of  the  other  two, 
ACD,  DCB,  is  also  equal  to  two  right  angles :  therefore,  the 
sum  of  the  four  is  equal  to  four  right  angles. 

In  general,  if  any  number  of  straight  lines 
CA,  CB,  CD,  &c.  meet  in  a  point  C,  the  ^ 
sum  of  all  the  successive  angles  ACB,BCD, 
DCE,  ECF,  FCA,  will  be  equal  to  four 
right  angles :  for,  if  four  right  angles  were 
formed  about  the  point  C,by  two  lines  per- 
pendicular to  each  other,  the  same  space 
would  be  occupied  by  the  four  right  angles,  as  by  the  succes- 
sive angles  ACB,  BCD,  DCE,  ECF,  FCA. 


PROPOSITION  V.    THEOREM. 

If  two  triangles  have  two  sides  and  the  included  angle  of  the  one, 
equal  to  two  sides  and  the  included  angle  of  the  other,  each  to 
each,  the  two  triangles  will  he  equal. 

Let  the  side  ED  be  equal 
lo  the  side  BA,  the  side  DF 
to  the  side  AC,  and  the  an- 
gle D  to  the  angle  A ;  then 
will  the  triangle  EDF  be 

equal  to  the  triangle  BAC. 

^.  £  !F  S  C 

^  For,  these  triangles  may  be  so  applied  to  each  other,  that  they 
shall  exactly  coincide.  Let  the  triangle  EDF,  be  placed  upon 
the  triangle  BAC,  so  that  the  point  E  shall  fall  upon  B,  and  the 
side  ED  on  the  equal  side  BA ;  then,  since  the  angle  D  is  equal 
to  the  angle  A,  the  side  DF  will  take  the  direction  AC.     But 


BOOK  I.  rt 

DF  is  equal  to  AC  ;  therefore,  the  point  F  will  fall  on  C,  and 
the  third  side  EF,  will  coincide  with  the  third  side  BC  (Ax.  11.): 
therefore,  the  triangle  EDF  is  equal  to  the  triangle  BAG 
(Ax.  13.). 

Cor,  When  two  triangles  have  these  three  things  equal, 
namely,  the  side  ED=BA,  the  side  DF=AC,  and  the  angle 
D=A,  the  remaining  three  are  also  respectively  equal,  namely, 
the  side  EF=BC,  the  angle  E=B,  and  the  angle  F=C 


PROPOSITION  VI.    THEOREM. 

If  two  triangles  have  two  angles  and  the  included  side  of  the  one, 
equal  to  two  angles  and  the  included  side  of  the  other,  each  to 
each,  the  two  triangles  will  he  equal. 

Let  the  angle  E  be  equal 
to  the  angle  B,  the  angle  F 
to  the  angle  C,  and  the  in- 
cluded side  EF  to  the  in- 
cluded side  BC ;  then  will 

the  triangle  EDF  be  equal  

to  the  triangle  BAC.  ^  ^  ^  ^ 

For  to  apply  the  one  to  the  other,  let  the  side  EF  be  placed 
on  its  equal  BC,  the  point  E  falling  on  B,  and  the  point  F  on 
C;  then,  since  the  angle  Eis  equal  to  the  angle  B,  the  side  ED 
will  take  the  direction  BA ;  and  hence  the  point  D  will  be  found 
somewhere  in  the  line  BA.  In  like  manner,  since  the  angle 
F  is  equal  to  Jhe  angle  C,  the  line  FD  will  take  the  direction 
CA,  and  the  point  D  will  be  found  somewhere  in  the  line  CA. 
Hence,  the  point  D,  falling  at  the  same  time  in  the  two  straight 
lines  B A  and  C A,  must  fall  at  their  intersection  A :  hence,  the 
two  triangles  EDF,  BAC,  coincide  with  each  other,  and  are 
therefore  equal  (Ax.  13.). 

Cor,  Whenever,in  two  triangles,these  three  things  are  equal, 
namely,  the  angle  E=B,  the  angle  F=C,  and  the  included  side 
EF  equal  to  the  included  side  BC,  it  may  be  inferred  that  the 
remaining  three  are  also  respectively  equal,  namely,  the  angle 
D=A,  the  side  ED=BA,  and  the  side  DF=AC. 

Scholium.  Two  triangles  are  said  to  be  equal,  when  being 
applied  to  each  other,  they  will  exactly  coincide  (Ax.  13.). 
Hence,  equal  triangles  have  their  like  parts  equal,  each  to  each, 
since  those  parts  must  coincide  with  each  other.  The  converse 
of  this  proposition  is  also  true,  namely,  that  two  triangles  which 
have  all  the  parts  of  the  one  equal  to  the  parts  of  the  other,  each 

B*3 


18 


GEOMETRY. 


to  eachf  are  equal ;  for  they  may  be  applied  to  each  other,  and 
the  equal  parts  will  mutually  coincide. 

PROPOSITION  VII.    THEOREM. 

The  sum  of  any  two  sides  of  a  triangle,  is  greater  than  the 
third  side. 

Let  ABC  be  a  triangle :  then  will  the 
sum  of  two  of  its  sides,  as  AC,  CB,  be 
greater  than  the  third  side  AB. 

For,  the  line  AB  is  the  shortest  dis- 
tance between  the  points  A  and  B 
<Def.  3.)  ;  hence  AC  +  CB  is  greater 
than  AB. 


PROPOSITION  VIII.     THEOREM. 

If  from  any  point  within  a  triangle,  two  straight  lines  he  drawn 
to  the  extremities  of  either  side,  their  sum  will  he  less  than  the 
^um  of  the  two  other  sides  of  the  triangle. 

Let  any  point,  as  O,  be  taken  within  the  trian- 
gle BAC,  and  let  the  lines  OB,  OC,  be  drawn 
to  the  extremities  of  either  side,  as  BC  ;  then 
willOB  +  OC<BA+AC. 

Let  BO  be  produced  till  it  meets  the  side  AC 
in  D  :  then  the  line  OC  is  shorter  than  OD  +  DC^ 
<Prop.  VIL):  add  BO  to  each,  and  we  have  BO-f  OC<BO  + 
OD  +  DC  (Ax.  4.),  or  BO  +  OC<BD  +  DC. 

Again,  BD<BA+ AD:  add  DC  to  each,  and  we  have  BD  + 
I)C<BA  + AC.  But  it  has  just  been  found  that  BO  +  OC< 
BD-f  DC  ;  therefore,  still  more  isB0  +  OC<BA  +  AC. 

PROPOSITION    IX.    THEOREM. 

If  two  triangles  have  two  sides  of  the  one  equal  to  two  sides  of  the 
other,  each  to  each,  and  the  included  angles  unequal,  the  third 
sides  will  he  unequal;  and  the  greater  side  will  helong  to  the 
triangle  which  has  the  greater  included  angle. 

Let  BAC  and  EDF 
be  two  triangles,  having 
the  sideAB=DE,  AC 
=DF,  and  the  angle 
A>D ;  then  will  BC> 
EF. 

Make  the  angle  CAGB 
=B;   take    AG=DE, 
and  draw  CG,       The 


BOOK  I. 


m 


triangle  GAC  is  equal  to  DEF,  since,  by  construction,  they 
have  an  equal  angle  in  each,  contained  by  equal  sides,  (Prop. 
V.) ;  therefore  CG  is  equal  to  EF.  Now,  there  may  be  three 
cases  in  the  proposition,  according  as  the  point  G  falls  without 
the  triangle  ABC,  or  upon  its  base  BC,  or  within  it. 

First  Case.  The  straight  hne  GC<GI  +  IC,  and  the  straight 
iine  AB<AI  +  IB;  therefore,  GC  +  AB<  GI+AI  +  IC  +  IB, 
or,  which  is  the  same  thing,  GC  +  AB<AG+BC.  Take  away 
AB  from  the  one  side,  and  its  equal  AG  from  the  other ;  and 
there  remains  GC<BC  (Ax.  5.)  ;  but  we  have  found  GC=EF, 
therefore,  BOEF. 


Second  Case.  If  the  point  G 
fall  on  the  side  BC,  it  is  evident 
that  GC,  or  its  equal  EF,  will  be 
shorter  than  BC  (Ax.  8.). 

B  & 

Third  Case,  Lastly,  if  the  point  G 
fall  within  the  triangle  BAC,  we  shall 
have,  by  the  preceding  theorem,  AG4- 
GC<AB  +  BC;  and,  taking  AG  from 
the  one,  and  its  equal  AB  from  the  other, 
there  will  remain  GC  <  BC  or  BC  >EF,  B 


Scholium.  Conversely,  if  two  sides 
BA,  AC,  of  the  triangle  BAC,  are  equal 
to  the  two  ED,  DF,of  the  triangle  EDF, 
each  to  each,  while  the  third  side  BC  of 
the  first  triangle  is  greater  than  the  third 
side  EF  of  the  second  ;  then  will  the  an- 
gle BAC  of  the  first  triangle,  be  greater 
than  the  angle  EDF  of  the  second. 

For,  if  not,  the  angle  BAC  must  be  equal  to  EDF,  or  less 
than  it.  In  the  first  case,  the  side  BC  would  be  equal  to  EF, 
(Prop.  V.  Cor.)  ;  in  the  second,  CB  would  be  less  than  EF  ;  but 
eitherof  these  results  contradicts  the  hypothesis :  therefore,  BAC 
is  greater  than  EDF, 


PROPOSITION  X.    THEOREM. 


/f  two  triangles  have  the  three  sides  of  the  one  equal  to  the  three 
sides  of  the  other^  each  to  each,  the  three  angles  will  also  he 
equal,  each  to  each,  and  the  triangles  themselves  will  he  equal. 


JIO  GEOMETRY. 


Let  the  side  ED=BA, 
the  side  EF=BC,  and  the 
side  DF=AC  ;  then  will 
the  angle  D=A,  the  angle 

E=B,   and  the   angle   F 

=  C.  E  TB  C 

For,  if  the  angle  D  were  greater  than  A,  while  the  sides 
ED,  DF,  were  equal  to  BA,  AC,  each  to  each,  it  would  fol- 
low, by  the  last  proposition,  that  the  side  EF  must  be  greater 
than  BC  ;  and  if  the  angle  t>  were  less  than  A,  it  would  follow, 
that  the  side  EF  must  be  less  than  BC  :  but  EF  is  equal  to  BC, 
by  hypothesis ;  therefore,  the  angle  D  can  neither  be  greater 
nor  less  than  A  ;  therefore  it  must  be  equal  to  it.  In  the  same 
manner  it  may  be  shown  that  the  angle  E  is  equal  to  B,  and 
the  angle  F  to  C :  hence  the  two  triangles  are  equal  (Prop. 
VI.  Sch.). 

Scholium.  It  may  be  observed  that  the  equal  angles  lie  op- 
posite the  equal  sides :  thus,  the  equal  angles  D  and  A,  He  op- 
posite the  equal  sides  EF  and  BC. 


PROPOSITION  XI.    THEOREM. 

In  an  isosceles  triangle,  the  angles  opposite  the  equal  sides 
are  equal. 

I>et  the  side  BA  be  equal  to  the  side  AC ;  then 
will  the  angle  C  be  equal  to  the  angle  B. 

For,  join  the  vertex  A,  and  D  the  middle  point 
of  the  base  BC.  Then,  the  triangles  BAD,  DAC, 
will  have  all  the  sides  of  the  one  equal  to  those 
of  the  other,  each  to  each  ;  for  BA  is  equal  to  AC,^" 
by  hypothesis ;  AD  is  common,  and  BD  is  equal 
to  DC  by  construction :  therefore,  by  the  last  proposition,  the 
angle  B  is  equal  to  the  angle  C. 

Cor.  An  equilateral  triangle  is  likewise  equiangular,  that  is 
to  say,  has  all  its  angles  equal.  * 

Scholium.  The  equality  of  the  triangles  BAD,  DAC,  proves 
also  that  the  angle  BAD,  is  equal  to  DAC,  and  BDA  to  ADC, 
hence  the  latter  two  are  right  angles ;  therefore,  the  line  drawn 
from  the  vertex  of  an  isosceles  triangle  to  the  middle  point  of  its 
base,  is  perpendicular  to  the  base,  and  divides  the  angle  at  the 
vertex  into  two  equal  parts. 

In  a  triangle  which  is  not  isosceles,  any  side  may  be  assumed 
indifferently  as  the  base ;  and  the  vertex  is,  in  that  case,  the 
vertex  of  the  opposite  angle.  In  an  isosceles  triangle,  however, 


BOOK  I.  Si 

that  side  is  generally  assumed  as  the  base,  which  is  not  equal 
to  either  of  the  other  two. 


PROPOSITION  XII.    THEOREM. 

Conversely,  if  two  angles  of  a  triangle  are  equal,  the  sides  oppo' 
site  them  are  also  equal,  and  the  triangle  is  isosceles. 

Let  the  angle  ABC  be  equal  to  the  angle  ACB ; 
then  will  the  side  AC  be  equal  to  the  side  AB. 

For,  if  these  sides  are  not  equal,  suppose  AB 
to  be  the  greater.  Then,  take  BD  equal  to  AC, 
and  draw  CD.  Now,  in  the  two  triangles  BDC, 
BAC,  we  have  BD=AC,  by  construction  ;  the 
angle  B  equal  to  the  angle  ACB,  by  hypothesis  ',]^J. 
and  the  side  BC  common  :  therefore,  the  two 
triangles,  BDC,  BAC,  have  two  sides  and  the  included  angle  in 
the  one,  equal  to  two  sides  and  the  included  angle  in  the  other, 
each  to  each :  hence  they  are  equal  (Prop.  V.).  But  the  part 
cannot  be  equal  to  the  whole  (Ax.  8.) ;  hence,  there  is  no 
inequality  between  the  sides  BA,  AC  ;  therefore,  the  triangle 
BAC  is  isosceles. 


PROPOSITION  XIII.    THEOREM. 

The  greater  side  of  everi^  triangle  is  opposite  to  the  greater  an- 
gle ;  and  conversely,  the  greater  angle  is  opposite  to  the 
greater  side. 

First,  hei  the  angle  C  be  greater  than  the  angle 
B  ;  then  will  the  side  AB,  opposite  C,  be  greater 
than  AC,  opposite  B. 

For,  make  the  angle  BCD=B.     Then,  in  the 
triangle  CDB,  we  shall  haveCD=BD  (Prop.  XII.). 
Now,  the  side  AC < AD  +  CD;  butAD+CD=C' 
AD  +  DB=AB:  therefore  AC<AB. 

Secondly,  Suppose  the  side  AB>AC;  then  will  the  angle  C, 
opposite  to  AB,  be  greater  than  the  angle  B,  opposite  to  AC. 

For,  if  the  angle  C<B,  it  follows,  from  what  has  just  been 
proved,  that  AB<AC;  which  is  contrary  to  the  hypothesis.  If 
the  angle  C=B,  then  the  side  AB=AC  (Prop.  XII.);  which  is 
also  contrary  to  the  supposition.  Therefore,  when  AB>AC, 
the  angle  C  must  be  greater  than  B. 


22  GEOMETRY. 


PROPOSITION  XIV.    THEOREM. 

From  a  given  pointy  without  a  straight  line,  only  one  perpendicu- 
lar can  he  drawn  to  that  line. 

Let  A  be  the  point,  and  DE  the  given 
line. 

Let  us  suppose  that  we  can  draw  two 
perpendiculars,  AB,  AC.  Produce  either 
of  them,  as  AB,  till  BF  is  equal  to  AB,  and  D^ 
draw  FC.  Then,  the  two  triangles  CAB, 
CBF,  will  be  equal:  for,  the  angles  CBA, 
and  CBF  are  right  angles,  the  side  CB  is  ^I* 

common,  and  the  side  AB  equal  to  BF,  by  construction  ;  there- 
fore, the  triangles  are  equal,  and  the  angle  ACB=BCF  (Prop. 
V.  Cor.).  But  the  angle  ACB  is  a  right  angle,  by  hypothesis  ; 
therefore,  BCF  must  likewise  be  a  right  angle.  But  if  the  adja- 
cent angles  BCA,  BCF,  are  together  equal  to  two  right  angles, 
ACF  must  be  a  straight  line  (Prop.  IIL) :  from  whence  it  fol- 
lows, that  between  the  same  two  points,  A  and  F,  two  straight 
lines  can  be  drawn,  which  is  impossible  (Ax.  11.)  J  hence,  two 
perpendiculars  cannot  be  drawn  from  the  same  point  to  the 
same  straight  line. 

Scholium.     At  a  given  point  C,  in  the  line  ^1 

AB,  it  is  equally  impossible  to  erect  two  per- 
pendiculars to  that  line.  For,  if  CD,  CE, 
were  those  two  perpendiculars,  the  angles 
BCD,  BCE,  would  both  be  right  angles:— 
hence  they  would  be  equal  (Ax.  10.);  and 
the  line  CD  would  coincide  with  CE ;  otherwise,  a  part  would 
be  equal  to  the  whole,  which  is  impossible  (Ax.  8.). 


PROPOSITION  XV.    THEOREM. 

If  from  a  point  without  a  straight  line,  a  perpendicular  he  let 
fall  on  the  line,  and  ohlique  lines  he  drawn  to  different  points : 

1st,  The  perpendicular  ivill  be  shorter  than  any  ohlique  line. 

2d,  Any  tioo  ohlique  lines,  drawn  on  different  sides  of  the  perpen- 
dicular, cutting  off  equal  distances  on  the  other  line,  will  he 
equal. 

2d,  Of  two  ohlique  lines,  drawn  at  pleasure,  that  which  is  farther 
from  the  peipendicular  will  he  the  longer. 


BOOK  I.  23 

Let  A  be  the  given  point,  DE  the  given 
line,  AB  the  perpendicular,  and  AD,  AC, 
AE,  the  oblique  lines. 

Produce  the  perpendicular  AB  till  BF 
is  equal  to  AB,  and  draw  FC,  FD.  D'" 

First.  The  triangle  BCF,  is  equal  to  the 
triangle  BCA,  for  they  have  the  right  angle 
CBF=CBA,  the  side  CB  common,  and  the  "^-^ 

side  BF=BA  ;  hence  the  third  sides,  CF  and  CA  are  equal 
(Prop.  V.  Cor.).  But  ABF,  being  a  straight  line,  is  shorter  than 
ACF,  v^^hich  is  a  broken  line  (Def.  3.)  ;  therefore,  AB,  the  half 
of  ABF,  is  shorter  than  AC,  the  half  of  ACF ;  hence,  the  per- 
pendicular is  shorter  than  any  obhque  line. 

Secondly.  Let  us  suppose  BC=BE;  then  will  the  triangle 
CAB  be  equal  to  the  the  triangle  BAE ;  for  BC=BE,the  side 
AB  is  common,  and  the  angle  CBA=ABE  ;  hence  the  sides 
AC  and  AE  are  equal  (Prop.  V.  Cor.) :  therefore,  two  oblique, 
lines,  equally  distant  from  the  perpendicular,  are  equal. 

Thirdly.  In  the  triangle  DFA,  the  sum  of  the  lines  AC,  CF, 
is  less  than  the  sum  of  the  sides  AD,  DF  (Prop.  VIII.) ;  there- 
fore, AC,  the  half  of  the  line  ACF,  is  shorter  than  AD,  the  half 
of  the  line  ADF :  therefore,  the  obhque  line,  which  is  farther 
from  the  perpendicular,  is  longer  than  the  one  which  is  nearer. 

Cor.  \.  The  perpendicular  measures  the  shortest  distance 
of  a  point  from  a  line. 

Cor.  2.  From  the  same  point  to  the  same  straight  line,  only 
two  equal  straight  lines  can  be  drawn  \  for,  if  there  could  be 
more,  we  should  have  at  least  two  equal  oblique  lines  on  the 
same  side  of  the  perpendicular,  which  is  impossible. 


PROPOSITION  XVI.    THEOREM. 

If  from  the  middle  point  of  a  straight  line,  a  perpendicular  he 
drawn  to  this  line ; 

\st,  Every  point  of  the  perpendicular  will  he  equally  distant 
from  the  extremities  of  the  line, 

2d,  Every  point,  without  the  perpendicular,  will  he  unequally  dis- 
tant from  those  extremities. 


124 


GEOMETRY. 


Let  AB  be  the  given  straight  line,  C  the 
middle  point,  and  ECF  the  perpendicular. 

First,  Since  AC  =  CB,  the  two  oblique  lines 
AD,  DB,  are  equally  distant  from  the  perpen- 
dicular, and  therefore  equal  (Prop.  XV.).  So, 
likewise,  are  the  two  oblique  lines  AE,  EB,  the^^ 
two  AF,  FB,  and  so  on.  Therefore  every  point 
in  the  perpendicular  is  equally  distant  from  the 
extremities  A  and  B. 

Secondly,  Let  I  be  a  point  out  of  the  perpen- 
dicular. If  lA  and  IB  be  drawn,  one  of  these  lines  will  cut 
the  perpendicular  in  D ;  from  which,  drawing  DB,  we  shall 
have  DB=DA.  But  the  straight  line  IB  is  less  than  ID  +  DB, 
and  ID  +  DB=ID  +  DA-IA;  therefore,  IB<IA;  therefore, 
every  point  out  of  the  perpendicular,  is  unequally  distant  from 
the  extremities  A  and  B. 

Cor.  If  a  straight  line  have  two  points  D  and  F,  equally  dis- 
tant from  the  extremities  A  and  B,  it  will  be  perpendicular  to 
AB  at  the  middle  point  C. 


PROPOSITION  XVII.    THEOREM. 

If  two  right  angled  triangles  have  the  hypothenuse  and  a  side  of 
the  one,  equal  to  the  hypothenuse  and  a  side  of  the  other,  each  to 
each,  the  remaining  parts  will  also  he  equal,  each  to  each,  and 
the  triangles  themselves  will  he  equal. 

In  the  two  right  angled  * 
triangles  BA€,  EDF,  let  the 
hypothenuse  AC=DF,  and 
the  side  BA=ED:  then  will 
the  side  BC==EF,  the  angle  ^ 
A=D,  and  the  angle  C=F. 

If  the  side  BC  is  equal  to  EF,  the  like  angles  of  the  two 
triangles  are  equal  (Prop.  X.).  Now,  if  it  be  possible,  suppose 
these  two  sides  to  be  unequal,  and  that  BC  is  the  greater. 

On  BC  take  BG=EF,  and  draw  AG.  Then,  in  the  two 
triangles  BAG,  DEF,  the  angles  B  and  E  are  equal,  being  right 
angles,  the  side  BA=:ED  by  hypothesis,  and  the  side  BG=EF 
by  construction  :  consequently,  AG =DF  (Prop.  V.  Cor.).  But, 
by  hypothesis  AC=DF;  and  therefore,  AC=AG  (Ax.  1.). 
But  the  oblique  line  AC  cannot  be  equal  to  AG,  which  lies 
nearer  the  perpendicular  AB  (Prop.  XV.) ;  therefore,  BC  and 
EF  cannot  be  unequal,  and  hence  the  angle  A=D,  and  the 
angle  C=F;  and  therefore,  the  triangles  are  equal  (Prop.  VI. 
Sch.).  ^       ^      f 


i 


BOOK  L  25 


PROPC^SITION  XVIII.    THEOREM. 

If  two  straight  lines  are  perpendicular  to  a  third  line,  they  will 
he  parallel  to  each  other :  in  other  words,  they  will  never  meet, 
how  far  soever  either  way,  both  of  them  he  produced. 

Let  the  two  lines  AC,  BD,  A  .  C 

be  perpendicular  to  AB  ;  then 
will  they  be  parallel. 

For,  iif  they  could  meet  in 
a  point  O,  on  either  side  of 
AB,  there  would  be  two  per- 


:r:::-0 


D 


pendiculars  OA,  OB,  let  fall  from  the  same  point  on  the  same 
straight  line;  which  is  impossible  (Prop.  XIV.). 


PROPOSITION  XIX.    THEOREM. 

If  two  straight  lines  meet  a  third  line,  making  the  sum  of  the 
interior  angles  on  the  same  side  of  the  line  met,  equal  to  two 
right  angles,  the  two  lines  will  he  parallel. 

Let  the  two  hues  EC,  BD,  meet 

the  third  line  BA,  making  the  an-  ^      A. __q 

gles  BAC,  ABD,  together  equal  to  j     / 

two  right  angles:  then  the  lines  r*!/ 

EC,  BD,  will  be  parallel.  y\ 

From  G,  the  middle  point  of       /  \ 

BA,  draw  the  straight  line  EGF,  -^^ — ^ :_ 

perpendicular  to  EC.     It  will  also    ^        *  ^^ 

be  perpendicular  to  BD.  For,  the  sum  BAC  +  ABD  is  equal 
to  two  right  angles,  by  hypothesis ;  the  sum  BAC  +  BAE  is 
likewise  equal  to  two  right  angles  (Prop.  I.) ;  and  taking  away 
BAC  from  both,  there  will  remain  the  angle  ABD = BAE. 

Again,  the  angles  EGA,  BGF,  are  equal  (Prop.  IV.) ;  there- 
fore, the  triangles  EGA  and  BGF,  have  each  a  side  and  two 
adjacent  angles  equal ;  therefore,  they  are  themselves  equal, 
and  the  angle  GEA  is  equal  to  the  angle  GFB  (Prop.  VI.  Cor.) : 
but  GEA  is  a  right  angle  by  construction  ;  therefore,  GFB  is  a 
right  angle ;  hence  the  two  lines  EC,  BD,  are  perpendicular  to 
the  same  straight  line,  and  are  therefore  parallel  (Prop.  XVIII.) . 


26  GEOMETRY. 

Scholium.  When  two  parallel 
straight  lines  AB,  CD,  are  met  by  a 
third  line  FE,  the  angles  which  are 
formed  take  particular  names. 

Interior  angles  on  the  same  side,  are 
those  which  lie  within  the  parallels,  ^ 
and  on  the  same  side  of  the  secant 
line :  thus,  0GB,  GOD,  are  interior 
angles  on  the  same  side  ;  and  so  also 
are  the  the  angles  OGA,  GOC. 

Alternate  angles  lie  within  the  parallels,  and  on  different 
sides  of  the  secant  line:  AGO,  DOG,  are  alternate  angles ; 
and  so  also  are  the  angles  COG,  BGO. 

Alternate  exterior  angles  lie  without  the  parallels,  and  on  dif- 
ferent sides  of  the  secant  line :  EGB,  COF,  are  alternate  exte- 
rior angles ;  so  also,  are  the  angles  AGE,  FOD. 

Opposite  exterior  and  interior  angles  lie  on  the  same  side  of  the 
secant  line,  the  one  without  and  the  other  within  the  parallels, 
but  not  adjacent :  thus,  EGB,  GOD,  are  opposite  exterior  and 
interior  angles ;  and  so  also,  are  the  angles  AGE,  GOC. 

Cor.  1.  If  a  straight  line  EF,  meet  two  straight  lines  CD, 
AB,  making  the  alternate  angles  AGO,  GOD,  equal  to  each 
other,  the  two  lines  will  be  parallel.  For,  to  each  add  the  an- 
gle OGB;  we  shall  then  have,  AGO  +  0GB -GOD  +  0GB  : 
but  AGO  +  OGB  is  equal  to  two  right  angles  (Prop.  I.) ;  hence 
GOD  +  OGB  is  equal  to  two  right  angles :  therefore,  CD,  AB, 
are  parallel. 

Cor.  2.  If  a  straight  line  EF,  meet  two  straight  lines  CD, 
AB,  making  the  exterior  angle  EGB  equal  to  the  interior  and 
opposite  angle  GOD,  the  two  lines  will  be  parallel.  For,  to  each 
add  the  angle  OGB:  we  shall  then  have  EGB  +  OGB = GOD 
+  OGB  :  but  EGB  +  OGB  is  equal  to  two  right  angles  ;  hence, 
GOD  +  OGB  is  equal  to  two  right  angles ;  therefore,  CD,  AB, 
are  parallel. 

PROPOSITION  XX.    THEOREM. 

If  a  straight  line  meet  two  parallel  straight  lines,  the  sum  of  the 

interior  angles  on  the  same  side  will  he  equal  to  two  right  angles. 

Let  the  parallels  AB,CD,be 
met  by  the  secant  line  FE :  then 
will  OGB  +  GOD,  or  OGA  + 

GOCjbe  equal  to  two  right  an-,  A ^'^""--^^ B 

gles. 

For,  if  OGB  +  GOD  be  not 

equal  to  two  right  angles,  let     _^ 

IGH  be  drawn, makingthe  sum  C        y/O  jD 

OGH  +  GOD   equal    to    two  F 


BOOK  I.  ^7 

right  angles ;  then  IH  and  CD  will  be  parallel  (Prop.  XIX.), 
and  hence  we  shall  have  two  lines  GB,  GH,  drawn  through 
the  same  point  G  and  parallel  to  CD,  which  is  impossible  (Ax. 
12.):  hence,  GB  and  GH  should  coincide,  and  0GB  +  GOD  is 
equal  to  two  right  angles.  In  the  same  manner  it  may  be  proved 
that  OGA+GOC  is  equal  to  two  right  angles. 

Cor.  1.  If  0GB  is  a  right  angle,  GOD  will  be  a  right  angle 
also:  therefore,  every  straight  line  perpendicular  to  one  of  two 
parallels,  is  perpendicular  to  the  other. 

Cor.  2.  If  a  straight  line  meet  tw^o 
parallel  lines,  the  alternate  angles  will 
be  equal. 

Let  AB,  CD,  be  the  parallels,  and 
FE  the  secant  line.  The  sum  0GB  + 
GOD  is  equal  to  two  right  angles.  But  g- 
the  sum  OGB  +  OGA  is  also  equal  to 
two  right  angles  (Prop.  I.).     Taking  /^ 

from  isach,  the  angle  OGB,  and  there 
remains  OGA=GOD.  In  the  same  manner  we  may  prove  that 
GOC=OGB. 

Cor.  3.  If  a  straight  line  meet  two  parallel  lines,  the  oppo- 
site exterior  and  interior  angles  will  be  equal.  For,  the  sum 
OGB  +  GOD  is  equal  to  two  right  angles.  But  the  sum  OGB 
-f  EGB  is  also  equal  to  two  right  angles.  Taking  from  each  the 
angle  OGB,  and  there  remains  GOD=EGB.  In  the  same 
manner  we  may  prove  that  AGE  =:GOC. 

Cor.  4.  We  see  that  of  the  eight  angles  formed  by  a  line 
cutting  two  parallel  lines  obliquely,  the  four  acute  angles  are 
equal  to  each  other,  and  so  also  are  the  four  obtuse  angles. 

PROPOSITION  XXI.    THEOREM. 

If  a  straight  line  meet  two  other  straight  lines,  making  the  sum  of 
the  interior  angles  on  the  same  side  less  than  two  right  angles, 
the  two  lines  will  meet  if  sufficiently  produced. 

Let  the  line  EF  meet  the  two 
lines  CD,  IH,  making  the  sum 
of  the  interior  angles  OGH, 
GOD,  less  than  two  right  an- 
gles :  then  will  IH  and  CD 
meet  if  sufficiently  produced. 

For,  if  they  do  not  meet  they 
are  parallel  (Def.  12.).  But  they 
are  not  parallel,  for  if  they  were, 
the  sum  of  the  interior  angles  OGH,  GOD,  would  be  equal  to 
two  right  angles  (Prop.  XX.),  whereas  it  is  less  by  hypothesis : 
hence,  the  lines  IH,  CD,  are  not  parallel,  and  will  therefore 
meet  if  sufficiently  produced. 


28 


GEOMETRY. 


Cor,  It  is  evident  that  the  two  lines  IH,  CD,  will  meet  on 
that  side  of  EF  on  which  the  sum  of  the  two  angles  OGH, 
GOD,  is  less  than  two  right  angles. 


PROPOSITION  XXII.    THEOREM. 

Two  straight  lines  which  are  parallel  to  a  third  line,  are  parallel 
to  each  other. 


Let  CD  and  AB  be  parallel  to  the  third  line  EF ;  then  are 
they  parallel  to  each  other. 

Draw  PQR  perpendicular  to  EF,  and 

cutting  AB,  CD.     Since  AB  is  parallel  to 

EF,  PR  will  be  perpendicular  to  AB  (Prop.E 

XX.  Cor.  1.) ;  and  since  CD  is  parallel  to 

EF,  PR  will  for  a  like  reason  be  perpen-C 

dicular  to  CD.     Hence  AB  and  CD  are 

perpendicular  to   the  same   straight  line  ;-^ 
hence  they  are  parallel  (Prop.  XVIII.). 


PROPOSITION  XXIII.    THEOREM. 
Two  parallels  dre  every  where  equally  distant. 

Two  parallels  AB,  CD,  being  o 
given,  if  through  two  points  E  "" 
and  F,  assumed  at  pleasure,  the 
straight  lines  EG,  FH,  be  drawn 
perpendicular  to  ABjthese  straight  ^ 
lines  will  at  the  same  time   be 
perpendicular  to  CD  (Prop.  XX.  Cor.  1.) :  and  we  are  now  to 
show  that  they  will  be  equal  to  each  other. 

If  GF  be  drawn,  the  angles  GFE,  FGH,  considered  in  refer- 
ence to  the  parallels  AB,  CD,  will  be  alternate  angles,  and 
therefore  equal  to  each  other  (Prop.  XX.  Cor.  2.).  Also,  the 
straight  lines  EG,  FH,  being  perpendicular  to  the  same  straight 
line  AB,  are  parallel  (Prop.  XYIII.) ;  and  the  angles  EGF, 
GFH,  considered  in  reference  to  the  parallels  EG,  FH,  will  be 
alternate  angles,  and  therefore  equal.  Hence  the  two  trian- 
gles EFG,  FGH,  have  a  common  side,  and  two  adjacent  angles 
in  each  equal ;  hence  these  triangles  are  equal  (Prop.  VI.)  ; 
therefore,  the  side  EG,  which  measures  the  distance  of  the 
parallels  AB  and  CD  at  the  point  E,  is  equal  to  the  side  FH, 
which  measures  the  distance  of  the  same  parallels  at  the 
point  F, 


\ 


BOOK  I.  29 


PROPOSITION  XXIV.    THEOREM. 

If  two  angles  have  their  sides  parallel  and  lying  in  the  same  di- 
rection, the  two  angles  will  be  equal. 

Let  BAG  and  DEF  be  the  two  angles, 
having  AB  parallel  to  ED,  and  AC  to  EF ; 
then  w^ill  the  angles  be  equal. 

For,  produce  DE,  if  necessary,  till  it 

meets  AC  in  G.     Then,  since  EF  is  par- 

allelto  GG,  the  angle  DEF  is  equal  to  ^         ^  ^ 

DGC   (Prop.  XX.   Cor.   3.);    and   since 
DG  is  parallel  to  AB,  the  angle  DGC  is  equal  to  BAG  ;  hence, 
the  angle  DEF  is  equal  to  BAG  (Ax.  1.). 

Scholium.  The  restriction  of  this  proposition  to  the  case 
where  the  side  EF  Hes  in  the  same  direction  with  AC,  and  ED 
in  the  same  direction  with  AB,  is  necessary,  because  if  FE 
were  produced  towards  H,  the  angle  DEH  would  have  its  sides 
parallel  to  those  of  the  angle  BAG,  but  would  not  be  equal  to 
it.  In  that  case,  DEH  and  BAG  would  be  together  equal  to 
two  right  angles.  For,  DEH  +  DEF  is  equal  to  two  right  angles 
(Prop.  I.) ;  but  DEF  is  equal  to  BAG :  hence,  DEH  +  BAG  is 
equal  to  two  right  angles. 

PROPOSITION  XXV.    THEOREM. 

In  every  triangle  the  sum  of  the  three  angles  is  equal  to  two 
right  angles. 

Let  ABC  be  any  triangle  :  then  will  the  an- 
gle C  +  A  +  B  be  equal  to  two  right  angles. 

For,  produce  the  side  CA  towards  D,  and  at 
the  point  A,  draw  AE  parallel  to  BC.  Then, 
since  AE,  CB,  are  parallel,  and  GAD  cuts  them, 
the  exterior  angle  DAE  will  be  equal  to  its  inte-  C 
rior  opposite  one  ACB  (Prop.  XX.  Cor.  3.) ;  in  like  manner, 
since  AE,  CB,  are  parallel,  and  AB  cuts  them,  the  alternate 
angles  ABC,  BAE,  will  be  equal :  hence  the  three  angles  of 
the  triangle  ABC  make  up  the  same  sum  as  the  three  angles 
GAB,  BAE,  EAD  ;  hence,  the  sum  of  the  three  angles  is  equal 
to  two  right  angles  (Prop.  L). 

Cor.  1.  Two  angles  of  a  triangle  being  given,  or  merely 
their  sum,  the  third  will  be  found  by  subtracting  that  sum  from 
two  right  angles. 


A       D 


30  GEOMETRY. 

Cor.  2.  If  two  angles  of  one  triangle  are  respectively  equal 
to  two  angles  of  another,  the  third  angles  will  also  be  equal, 
and  the  two  triangles  will  be  mutually  equiangular. 

Cor.  3.  In  any  triangle  there  can  be  but  one  right  angle  ; 
for  if  there  were  two,  the  third  angle  must  be  nothing.  Still 
less,  can  a  triangle  have  more  than  one  obtuse  angle. 

Cor.  4.  In  every  right  angled  triangle,  the  sum  of  the  two 
acute  angles  is  equal  to  one  right  angle. 

Cor.  5.  Since  every  equilateral  triangle  js  also  equiangular 
(Prop.  XI.  Cor.),  each  of  its  angles  will  be  equal  to  the  third 
part  of  two  right  angles ;  so  that,  if  the  right  angle  is  expressed 
by  unity,  the  angle  of  an  equilateral  triangle  will  be  expressed 

Cor.  6.     In  €very  triangle  ABC,  the  exterior  angle  BAD  is 
equal  to  the  sum  of  the  two  interior  opposite  angles  B  and  C. 
J^         For,  AE  being  parallel  to  BC,  the  part  BAE  is  equal  to  the 
•^  angle  B,  and  the  other  part  DAE  is  equal  to  the  angle  C. 


PROPOSITION  XXVI.    THEOREM. 

The  sum  of  all  the  interior  angles  of  a  polygon,  is  equal  to  two 
right  angles,  taken  as  many  times  less  two,  as  the  figure  has 
sides. 

Let  ABCDEFG  be  the  proposed  polygon. 
If  from  the  vertex  of  any  one  angle  A,  diagonals  ^^ 
AC,  AD,  AE,  AF,  be  drawn  to  the  vertices  of 
all  the  opposite  angles,  it  is  plain  that  the  poly- 
gon will  be  divided  into  five  triangles,  if  it  has' 
seven  sides  ;  into  six  triangles,  if  it  has  eight;  and, 
in  general,  into  as  many  triangles,  less  two,  as 
the  polygon  has  sides ;  for,  these  triangles  may  be  considered 
as  having  the  point  A  for  a  common  vertex,  and  for  bases,  the 
several  sides  of  the  polygon,  excepting  the  two  sides  which  form 
the  angle  A.  It  is  evident,  also,  that  the  sum  of  all  the  angles 
in  these  triangles  does  not  difter  from  the  sum  of  all  the  angles 
in  the  polygon :  hence  the  sum  of  all  the  angles  of  the  polygon 
is  equal  to  two  right  angles,  taken  as  many  times  as  there  are 
triangles  in  the  figure  ;  in  other  words,  as  there  are  units  in  the 
number  of  sides  diminished  by  two. 

Cor.  1.     The  sum  of  the  angles  in  a  quadrilateral  is  equal 
to  two  right  angles  multiplied  by  4 — 2,  which  amounts  to  four 


BOOK  I.  31 

right  angles :  hence,  if  all  the  angles  of  a  quadrilateral  are 
equal,  each  of  them  will  be  a  right  angle  ;  a  conclusion  which 
sanctions  the  seventeenth  Definition,  where  the  four  angles  of 
a  quadrilateral  are  asserted  to  be  right  angles,  in  the  case  of  the 
rectangle  and  the  square. 

Cor,  2.  The  sum  of  the  angles  of  a  pentagon  is  equal  to 
two  right  angles  multiplied  by  5 — 2,  which  amounts  to  six  right 
angles :  hence,  when  a  pentagon  is  equiangular,  each  angle 
is  equal  to  the  fifth  part  of  six  right  angles,  or  to  |  of  one  right 
angle. 

Cor.  3.  The  sum  of  the  angles  of  a  hexagon  is  equal  to 
2  X  (6 — 2,)  or  eight  right  angles ;  hence  in  the  equiangular 
hexagon,  each  angle  is  the  sixth  part  of  eight  right  angles,  or  | 
of  one. 

Scholium.  When  this  proposition  is  appHed  to 
polygons  which  have  re-entrant  angles,  each  re- 
entrant angle  must  be  regarded  as  greater  than 
two  right  angles.  But  to  avoid  all  ambiguity,  we 
shall  henceforth  limit  our  reasoning  to  polygons 
with  salient  angles,  which  might  otherwise  be  named  convex 
polygons.  Every  convex  polygon  is  such  that  a  straight  line, 
drawn  at  pleasure,  cannot  meet  the  contour  of  the  polygon  in 
more  than  two  points. 


PROPOSITION  XXVII.    THEOREM. 

If  the  sides  of  any  polygon  be  produced  out,  in  the  same  direc- 
tion, the  sum^ofthe  exterior  angles  will  be  equal  to  four  right 
angles. 

Let  the  sides  of  the  polygon  ABCD- 
FG,  be  produced,  in  the  same  direction ; 
then  will  the  sum  of  the  exterior  angles 
a  +  b-{-c-\-d  +/+  gi  be  equal  to  four  right 
angles. 

For,  each  interior  angle,  plus  its  ex- 
terior angle,  as  A  +  a,  is  equal  to  two 
right  angles  (Prop.  I.).  But  there  are 
as  many  exterior  as  interior  angles,  and  as  many  of  each  as 
there  are  sides  of  the  polygon  :  hence,  the  sum  of  all  the  inte- 
rior and  exterior  angles  is  equal  to  twice  as  many  right  angles 
as  the  polygon  has  sides.  Again,  the  sum  of  all  the  interior 
angles  is  equal  to  two  right  angles,  taken  as  many  times,  less 
two,  as  the  polygon  has  sides  (Prop.  XXVI.) ;  that  is,  equal  to 
twice  a;j  many  right  angles  as  the  figure  has  sides,  wanting 
four  right  angles.     Hence,  the  interior  angles  plus  four  right 


32  GEOMETRY. 

angles,  is  equal  to  twice  as  many  right  angles  as  the  polygon 
has  sides,  and  consequently,  equal  to  the  sum  of  the  interior 
angles  plus  the  exterior  angles.  Taking  from  each  the  sum  of 
the  interior  angles,  and  there  remains  the  exterior  angles,  equal 
to  four  right  angles. 


PROPOSITION  XXVIII.     THEOREM. 

In  every  parallelogram,  the  opposite  sides  and  angles  are  equal. 

Let  ABC D  be  a  parallelogram:  then  will       d  c 

AB=DC,  AD=BC,  A=C,  and  ADC  =ABC. 

For,  draw  the  diagonal  BD.  The  triangles 
ABD,  DBC,  have  a  common  side  BD ;  and 
since  AD,  BC,  are  parallel,  they  have  also  the 


angle  ADB=DBC,  (Prop.  XX.  Cor.  2.) ;  and  since  AB,  CD, 
are  parallel,  the  angle  ABD=BDC :  hence  the  two  triangles 
are  equal  (Prop.  VI.) ;  therefore  the  side  AB,  opposite  the  an- 
gle ADB,  is  equal  to  the  side  DC,  opposite  the  equal  angle 
DBC  ;  and  the  third  sides  AD,  BC,  are  equal :  hence  the  op- 
posite sides  of  a  parallelogram  are  equal. 

Again,  since  the  triangles  are  equal,  it  follows  that  the  angle 
A  is  equal  to  the  angle  C  ;  and  also  that  the  angle  ADC  com- 
posed of  the  two  ADB,  BDC,  is  equal  to  ABC,  composed  of 
the  two  equal  angles  DBC,  ABD  :  hence  the  opposite  angles 
of  a  parallelogram  are  also  equal. 

Cor.  Two  parallels  AB,  CD,  included  between  two  other 
parallels  AD,  BC,  are  equal ;  and  the  diagonal  DB  divides  the 
parallelogram  into  two  equal  triangles. 


PROPOSITION  XXIX.    THEOREM. 

If  the  opposite  sides  of  a  quadrilateral  are  equal,  each  to  each, 
the  equal  sides  will  be  parallel,  and  the  figure  will  he  a  par- 
allelogram. 

Let  ABCD  be  a  quadrilateral,  having 
its  opposite  sides  respectively  equal,  viz. 
AB-=DC,  and  AD=BC  ;  then  will  these 
sides  be  parallel,  and  the  figure  be  a  par- 
allelogram. 

For,  having  drawn  the  diagonal  BD, 
the  triangles  ABD,  BDC,  have  all  the  sides  of  the  one  equal  to 


BOOK  I.  33 

the  corresponding  sides  of  the  other ;  therefore  they  are  equal, 
and  the  angle  ADB,  opposite  the  side  AB,  is  equal  to  DBC, 
opposite  CD  (Prop.  X.) ;  therefore,  the  side  AD  is  parallel  to 
BC  (Prop.  XIX.  Cor.  1.).  For  a  like  reason  AB  is  parallel  to 
CD  :  therefore  the  quadrilateral  ABCD  is  a  parallelogram. 

PROPOSITION  XXX.    THEOREM. 

If  two  opposite  sides  of  a  quadrilateral  are  equal  and  parallel, 
the  remaining  sides  will  also  be  equal  and  parallel,  and  the 
figure  will  be  a  parallelogram. 

Let  ABCD  be  a  quadrilateral,  having 
the  sides  AB,  CD,  equal  and  parallel ; 
then  will  the  figure  be  a  parallelogram. 

For,  draw  the  diagonal  DB,  dividing 
the  quadrilateral  into  two  triangles.  Then, 
since  AB  is  parallel  to  DC,  the  alternate 
angles  ABD,  BDC,  are  equal  (Prop.  XX.  Cor.  2.) ;  moreover, 
the  side  DB  is  common,  and  the  side  AB=DC  ;  hence  the  tri- 
angle ABD  is  equal  to  the  triangle  DBC  (Prop.  V.) ;  therefore, 
the  side  AD  is  equal  to  BC,  the  angle  ADB  =  DBC,  and  conse- 
quently AD  is  parallel  to  BC ;  hence  the  figure  ABCD  is  a 
parallelogram. 

PROPOSITION  XXXI.    THEOREM. 

The  two  diagonals  of  a  parallelogram  divide  each  other  into  equal 
parts,  or  mutually  bisect  each  other. 

Let  ABCD  be  a  parallelogram,  AC  and  3 
DB  its  diagonals,  intersecting  at  E,then  will 
AE=EC,  and  DE=EB. 

Comparing  the  triangles  ADE,  CEB,  we 
find  the  side  AD=CB  (Prop.  XXVIIL), 
the  angle  ADE  =  CBE,  and  the  angle 
DAE=ECB  (Prop.  XX.  Cor.  2.) ;  hence  those  triangles  are 
equal  (Prop.  VI.) ;  hence,  AE,  the  side  opposite  the  angle 
ADE,  is  equal  to  EC,  opposite  EBC  ;  hence  also  DE  is  equal 
toEB. 

Scholium.  In  the  case  of  the  rhombus,  the  sides  AB,  BC, 
being  equal,  the  triangles  AEB,  EBC,  have  all  the  sides  of  the 
one  equal  to  the  corresponding  sides  of  the  other,  and  are 
therefore  equal :  whence  it  follows  that  the  angles  AEB,  BEC, 
are  eqtial,  and  therefore,  that  the  two  diagonals  of  a  rhombus 
cut  each  other  at  right  angles. 

5 


34  GEOMETRY. 

BOOK  n. 

OF  RATIOS   AND   PROPORTIONS. 

Definitions. 

1.  Ratio  is  the  quotient  arising  fram  dividing  one  quantity 
by  another  quantity  of  the  same  kind.  Thus,  if  A  and  B  rep- 
resent quantities  of  the  same  kind,  the  ratio  of  A  to  B  is  ex- 
pressed by  -^. 

The  ratios  of  magnitudes  may  be  expressed  by  numbers, 
either  exactly  or  approximatively ;  and  in  the  latter  case,  the 
approximation  may  be  brought  nearer  to  the  true  ratio  than 
any  assignable  difference. 

Thus,  of  two  magnitudes,  one  of  them  may  be  considered  to 
be  divided  into  some  number  of  equal  parts,  each  of  the  same 
kind  as  the  whole,  and  one  of  those  parts  being  considered  as 
an  unit  of  measure,  the  magnitude  may  be  expressed  by  the 
number  of  units  it  contains.  If  the  other  magnitude  contain 
a  certain  number  of  those  units,  it  also  may  be  expressed  by 
the  number  of  its  units,  and  the  two  quantities  are  then  said 
to  be  commensurable. 

If  the  second  magnitude  do  not  contain  the  measuring  unit 
an  exact  number  of  times,  there  may  perhaps  be  a  smaller  unit 
which  will  be  contained  an  exact  number  of  times  in  each  of 
the  magnitudes.  But  if  there  is  no  unit  of  an  assignable  value, 
which  shall  be  contained  an  exact  number  of  times  in  each  of 
the  magnitudes,  the  magnitudes  are  said  to  be  incommensurable. 

It  is  plain,  however,  that  the  unit  of  measure,  repeated  as 
many  times  as  it  is  contained  in  the  second  magnitude,  would 
always  differ  from  the  second  magnitude  by  a  quantity  less 
than  the  unit  of  measure,  since  the  remainder  is  always  less 
than  the  divisor.  Now,  since  the  unit  of  measure  may  be  made 
as  small  as  we  please,  it  follows,  that  magnitudes  may  be  rep- 
resented by  numbers  to  any  degree  of  exactness,  or  they  will 
differ  from  their  numerical  representatives  by  less  than  any 
assignable  quantity. 

Therefore,  of  two  magnitudes,  A  and  B,  we  may  conceive 
A  to  be  divided  into  M  number  of  units,  each  equal  to  A' : 
then  A=M  x  A':  let  B  be  divided  into  N  number  of  equal  units, 
each  equal  to  A';  then  B=N  x  A';  M  and  N  being  integral  num- 
bers. Now  the  ratio  of  A  to  B,  will  be  the  same  as  the  ratio 
of  M  X  A'  to  N  X  A';  that  is  the  same  as  the  ratio  of  M  to  N,  since 
A'  is  a  common  unit. 


BOOK  II.  35 

In  the  same  manner,  the  ratio  of  any  other  two  magnitudes 
C  and  D  may  be  expressed  by  P  x  C  to  Q  x  C,  P  and  Q  being 
also  integral  numbers,  and  their  ratio  will  be  the  same  as  that 
ofPtoQ. 

2.  If  there  be  four  magnitudes  A,  B,  C,  and  D,  having  such 

B  D 

values  that  —is  equal  to— ,  then  A  is  said  to  have  the  same  ratio 

A  "O 

to  B,  that  C  has  to  D,  or  the  ratio  of  A  to  B  is  equal  to  the  ratio 
of  C  to  D.  When  four  quantities  have  this  relation  to  each 
other,  they  are  said  to  be  in  proportion. 

To  indicate  that  the  ratio  of  A  to  B  is  equal  to  the  ratio  of 
C  to  D,  the  quantities  are  usually  written  thus,  A :  B  : :  C  :  D, 
and  read,  A  is  to  B  as  C  is  to  D.  The  quantities  which  are 
compared  together  are  called  the  terms  of  the  proportion.  The 
first  and  last  terms  are  called  the  two  extremes,  and  the  second 
and  third  terms,  the  two  means. 

3.  Of  four  proportional  quantities,  the  first  and  third  are 
called  the  antecedents,  and  the  second  and  fourth  the  conse- 
quents ;  and  the  last  is  said  to  be  a  fourth  proportional  to  the 
other  three  taken  in  order. 

4.  Three  quantities  are  in  proportion,  when  the  first  has  the 
same  ratio  to  the  second,  that  the  second  has  to  the  third  ;  and 
then  the  middle  term  is  said  to  be  a  mean  proportional  between 
the  other  two. 

5.  Magnitudes  are  said  to  be  in  proportion  by  inversion,  or 
inversely,  when  the  consequents  are  taken  as  antecedents,  and 
the  antecedents  as  consequents. 

6.  Magnitudes  are  in  proportion  by  alternation,  cr  alternately, 
when  antecedent  is  compared  with  antecedent,  and  consequent 
with  consequent. 

7.  Magnitudes  are  in  proportion  by  composition,  when  the 
sum  of  the  antecedent  and  consequent  is  compared  either  with 
antecedent  or  consequent. 

8.  Magnitudes  are  said  to  be  in  proportion  by  division,  when 
the  difference  of  the  antecedent  and  consequent  is  compared 
either  with  antecedent  or  consequent. 

9.  Equimultiples  of  two  quantities  are  the  products  which 
arise  from  multiplying  the  quantities  by  the  same  number : 
thus,  m  X  A,  m  X  B,  are  equimultiples  of  A  and  B,  the  common 
multipHer  being  m. 

10.  Two  quantities  A  and  B  are  said  to  be  reciprocally  pro- 
portional, or,  invasely  proportional,  when  one  of  them  is  pro- 
portional to  unity  divided  by  the  other,  in  which  case  their  pro- 
duct is  always  equal  to  a  constant  quantity. 


GEOMETRY. 


PROPOSITION  I,    THEOREM. 


When  four  quantities  are  in  proportion^  the  product  of  the 
two  extremes  is  equal  to  the  product  of  the  two  means. 

Let  A,  B,  C,  D,  be  four  quantities  in  proportion,  and  M  :  N 
: :  P  :  Q  be  their  numerical  representatives ;  then  will  M  x  Q= 

N    O 
Nx  P;  for  since  the  quantities  are  in  proportion  :jr;i:=5  there- 
fore N=Mxpi,orNxP=MxQ. 

Cor.  If  there  are  three  proportional  quantities  (Def.  4.),  the 
product  of  the  extremes  will  be  equal  to  the  square  of  the 
mean. 


PROPOSITION  II.     THEOREM. 

If  the  'product  of  two  quantities  he  equal  to  the  product  of  two  other 
quantities,  two  of  them  will  he  the  extremes  and  the  other  two 
the  means  of  a  proportion. 

Let  M  X  Q =N  X  P  ;  then  will  M :  N  : :  P  :  Q. 

For,  if  P  have  not  to  Q  the  ratio  which  M  has  to  N,  let  P 
have  to  Q',  a  number  greater  or  less  than  Q,  the  same  ratio 
that  M  has  to  N ;  that  is,  let  M  :  N  : :  P :  Q';  then  M  x  N=P  x 

Q'(Prop.L):  hence,  Q'=M^;  but  Q=:M^  ;  conse- 
quently Q=Q',  and  the  four  quantities  are  proportional ;  that 
isM:N;:P:Q. 


PROPOSITION  III.    THEOREM. 

If  four  quantities  are  in  proportion,  they  will  he  in  proportion 
when  taken  alternately. 

Let  M,  N,  P,  Q,  be  the  numerical  representatives  of  four 
quanties  in  proportion  ;  so  that 

M  :  N  : :  P  :  Q,  then  will  M  :  P  : :  N  :Q. 

Since  M  :  N  : :  P  :  Q,  by  supposition,  M  x  Q=N  x  P  ;  there- 
fore, M  and  Q  may  be  made  the  extremes,  and  N  and  P  the 
means  of  a  proportion  (Prop.  IL) ;  hence,  M  :  P  : :  N  :  Q. 


BOOK  11.  37 


PROPOSITION  IV.    THEOREM. 

If  there  he  four  proportional  quantities,  and  four  other  propor- 
tional quantities,  having  the  antecedents  the  same  in  both,  the 
consequents  will  he  proportional. 

Let  M  :  N  : :  P  :  Q 

and  M:R::P:S 

then  will  N :  Q  : :  R  :  S 

P     O 
For,  by  alternation      M :  P : :  N :  Q,  or     51 =^ 

P      S 
and  M  :  P  : :  R  ;  S,  or     jjj=g* 

Q     S 
hence  n'~R  '  ^^  N  :  Q : ;  R :  S. 

Cor,  If  there  be  two  sets  of  proportionals,  having  an  ante- 
cedent and  consequent  of  the  first,  equal  to  an  antecedent  and 
consequent  of  the  second,  the  remaining  terms  will  be  propor- 
tional. 


PROPOSITION  V.    THEOREM. 

If  two  quantities  he  in  proportion,they  will  he  in  proportion  when 
taken  inversely. 

Let  M  :  N  : :  P  :  Q, ;  then  will 

N  :  M  : :  Q  ;  P. 
For,  from  the  first  proportion  we  have  M  x  Q=N  x  P,  or 
NxP=MxQ. 

But  the  products  N  x  P  and  M  x  Q  are  the  products  of  the 
extremes  and  means  of  the  four  quantities  N,  M,  Q,  P,  and  these 
products  being  equal, 

N:M::Q:P(Prop.IL). 


PROPOSITION  VI.    THEOREM. 

If  four  quantities  are  in  proportion,  they  will  he  in  proportion  by 
^  composition,  or  division. 


D 


38  GEOMETRY. 

Let,  as  before,  M,  N,  P,  Q,  be  the  numerical  representative* 
of  the  four  quantities,  so  that 

M  :  N  : :  P  :  Q  ;  then  will 
M=bN:M::P±Q:P. 
For,  from  the  first  proportion,  we  have 

MxQ=NxP,  orNxP=MxQ; 
Add  each  of  the  members  of  the  last  equation  to,  or  subtract 
it  from  M.P,  and  we  shall  have, 

M.P±N.P=M.P±M.Q;  or 
(MdbN)xP==(P±Q)xM. 
But  MdzN  and  P,  may  be  considered  the  two  extremes,  and 
PdbQ  and  M,  the  two  means  of  a  proportion ;  hence, 

M±N  ;  M  : :  P±Q  :  P. 

PROPOSITION  VII.    THEOREM. 

Equimultiples  of  any  two  quantities,  have  the  same  ratio  as  the 
quantities  themselves. 

Let  M  and  N  be  any  two  quantities,  and  m  any  integral 
number ;  then  will 

m.  M  :  m.  N  : :  M  :  N.     For 

m.  MxN=77i.  NxM,  since  the  quantities  in 
each  member  are  the  same ;  therefore,  the  quantities  are  pro- 
portional (Prop.  II.) ;  or 

m.  M  :  m.  N  : :  M  :  N. 

PROPOSITION  VIII.    THEOREM. 

Of  four  proportional  quantities,  if  there  he  taken  any  equimul- 
tiples of  the  two  antecedents,  and  any  equimultiples  of  the  two 
consequents,  the  four  resulting  quantities  ivill  be  proportional. 

Let  M,  N,  P,  Q,  be  the  numerical  representatives  of  four 
quantities  in  proportion ;  and  let  m  and  n  be  any  numbers 
whatever,  then  will 

771.  M  :  ?i.  N  : :  m.  P  :  ?i.  Q. 

For,  since  M  :  N  : :  P  :  Q,  we  have  M  x  Q=N  x  P ;  hence, 
m.  Mxn.  Q=7i.  Nxm.  P,  by  multiplying  both  members  of  the 
equation  hy  mxn.  But  m.  M  and  n.  Q,  may  be  regarded  as 
the  two  extremes,  and  n.  N  and  m.  P,  as  the  means  of  a  propor- 
tion ;  hence,  m.Min.  N  : :  m.  P  :  w.  Q. 


BOOK  II.  39 


PROPOSITION  IX.    THEOREM. 

Of  four  proportional  quantities^  if  the  two  consequents  he  either 
augmented  or  diminished  hy  quantities  which  have  the  same 
ratio  as  the  antecedents,  the  resulting  quantities  and  the  ante- 
cedents will  he  proportional. 


Let 

M  :  N  :  :  P  :  Q,  and  let  also 

M  :  P  :  :  m  :  w,  then  will 

M  :  P  :  :  N=fcm  :  Q±«. 

For,  since 

M  :  N  :  :  P  :  Q,  MxQ=NxP. 

And  since 

M  :  P  :  :  m  :  ?i,  Mx?i=Px7» 

Therefore, 

MxQ±Mxn=NxP±Px?w 

or, 

Mx(Q±7i)=Px(N±m): 

hence 

M  :  P  :  ;  N±7w  :  Q±7i  (Prop.  II.) 

PROPOSITION  X.    THEOREM. 

If  any  number  of  quantities  are  proportionals,  any  one  antece- 
dent wilt  be  to  its  consequent,  as  the  sum  of  all  the  antecedents 
to  the  sum  of  the  consequents. 

Let  M  :  N  :  :  P  :  Q  :  :  R  :  S,  &c.  then  will 

M  :  N  :  :  M-f-P  +  R  :  N  +  Q  +  S 
For,  since         M  :  N  :  :  P  :  Q,  we  have  MxQ=NxP 
And  since         M  :  N  :  :  R  :  S,  we  have  Mx  S=NxR 
Add  MxN=MxN 

and  we  have,    M.N  +  M.Q  +  M.S=M.N  +  N.P-fN.R 
or     Mx(Ni-Q  +  S)=Nx(M  +  P  +  R) 
therefore,  M  :  N  :  :  M  +  P  +  R  :  N  +  Q  +  S. 


PROPOSITION  XI.    THEOREM. 

Jf  tioo  magnitudes  he  each  increased  or  diminished  hy  like  parts 
of  each,  the  resulting  quantities  will  have  the  same  ratio  as  the 
magnitudes  themselves. 


40  GEOMETRY. 

M        N 
Let  M  and  N  be  any  two  magnitudes,  and —  and  —  be  like 

parts  of  each :  then  will 

M  :  N  :  :  M±M  :  N  ±?^ 
m  m 

For,  it  is  obvious  that  Mx{N±^\   =Nx(M±M.\  since 

m  f  m  / 

each  is  equal  to  M.N±— l—.     Consequently,  the  four  quan- 

m 

titles  are  proportional  (Prop.  II.). 


PROPOSITION  XII.    THEOREM. 

If  four  quantities  are  proportional,  their  squares  or  cubes  will 
also  he  proportional. 

Let  M  :  N  :  P  :  Q, 

then  will  M^  :  N^  :  :  P^  :  Q2 

and  M^  :  N3  :  :  P3  :  Q3 

Tor,      MxQ=NxP,  since  M  :  N  :  :  P  :  Q 

or,      M^  X  Q^=N'^  X  P^      by  squaring  both  numbers, 
and       M^  X  Q^=N3  X  P^      by  cubing  both  numbers  ; 
therefore,     M^  :  N^  :  :  P2  :  Q2 
and       M3  :  N=^  :  :  P3  :  Q3 

Cor,  In  the  same  way  it  may  be  shown  that  like  powers  or 
roots  of  proportional  quantities  are  proportionals. 


PROPOSITION  XIII.    THEOREM. 

If  there  he  two  sets  of  proportional  quantities,  the  products  of  the 
corresponding  terms  will  he  proportional. 


Let 

M  :  N  :  :  P  :  Q 

and 

R  :  S  :  :  T  :  V 

then  will 

MxR  :  NxS  :  :  PxT  :  QxV. 

For  since 

MxQ=NxP 

and 

R  X  V= S  X  T,  we  shall  have 

MxQxRxV=NxPxSxT 

or 

MxRxQxV=NxSxPxT 

therefore. 

MxR  :  NxS  :  :  PxT  :  QxV. 

BOOK  III.  41 

BOOK  III. 

THE  CIRCLE,  AND  THE  MEASUREMENT  OF  ANGLES. 

Definitions, 

1.  The  circumference  of  a  circle  is  a 
curved  line,  all  the  points  of  which  are 
equally  distant  from  a  point  within, 
called  the  centre. 

The  circle  is  the  space  terminated  by  A| 
this  curved  line.* 

2.  Every  straight  line,  CA,  CE,  CD, 
drawn  from  the  centre  to  the  circum- 
ference, is  called  a  radium  or  semidiam-  '^ 
eter ;  every  line  which,  hke  AB,  passes  through  the  centre,  and 
is  terminated  on  both  sides  by  the  circumference,  is  called  a 
diameter. 

From  the  definition  of  a  circle,  it  follows  that  all  the  radii 
are  equal ;  that  all  the  diameters  are  equal  also,  and  each 
double  of  the  radius. 

3.  A  portion  of  the  circumference,  such  as  FHG,  is  called 
aji  arc. 

The  chordy  or  subtense  of  an  arc,  is  the  straight  line  FG,  which 
joins  its  two  extremities.f 

4.  A  segment  is  the  surface  or  portion  of  a  circle,  included 
between  an  arc  and  its  chord. 

5.  A  sector  is  the  part  of  the  circle  included  between  an 
arc  DE,  and  the  two  radii  CD,  CE,  drawn  to  the  extremities 
of  the  arc. 

6.  A  straight  line  is  said  to  be  inscribed  in 
a  circle,  when  its  extremities  are  in  the  cir- 
cumference, as  AB. 

An  inscribed  angle  is  one  which,  like  BAC, 
haS  its  vertex  in  the  circumference,  and  is 
formed  by  two  chords. 


*  Note.  In  common  language,  the  circle  is  sometimes  confoun4ed  with  its 
circumference :  but  the  correct  expression  may  always  be  easily  recurred  to  if 
we  bear  in  mind  that  the  circle  is  a  surface  which  has  length  and  breadthj 
while  the  circumference  is  but  a  line. 

t  Note.  In  all  cases,  the  same  chord  FG  belongs  to  two  arcs,  FGH,  FEG, 
and  consequently  also  to  two  segments  :  but  the  smaller  one  is  always  meant 
u?iless  the  contrary  is  expressed. 

D*6 


4S  GEOMETRY. 

An  inscribed  triangle  is  one  which,  like  BAG,  has  its  three 
angular  points  in  the  circumference. 

And,  generally,  an  inscribed  figure  is  one,  of  which  all  the 
angles  have  their  vertices  in  the  circumference.  The  circle  is 
then  said  to  circumscribe  such  a  figure. 

7.  A  secant  is  a  line  which  meets  the  circum- 
ference in  two  points,  and  lies  partly  within ' 
and  partly  without  the  circle.   AB  is  a  secant. 

8.  A  tangent  is  a  line  which  has  but  one 
point  in  common  with  the  circumference.  GD 
is  a  tangent. 

The  point  M,  where  the  tangent  touches  the  ^ 
circumference,  is  called  the  point  of  contact. 


In  like  manner,  two  circumferences  touch 
each  other  when  they  have  but  one  point  in 
common. 


9.  A  polygon  is  circumscribed  about  a 
circle,  when  all  its  sides  are  tangents  to 
the  circumference  :  in  the  same  case,  the 
circle  is  said  to  be  inscribed  in  the  po- 
lygon. 


PROPOSITION  I.    THEOREM. 

Every  diameter  divides  the  circle  and  its  circumference  into  two 
equal  parts. 

Let  AEDF  be  a  circle,  and  AB  a  diameter. 
Now,  if  the  figure  AEB  be  applied  to  AFB, 
their  common  base  AB  retaining  its  position, 
the  curve  line  AEB  must  fall  exactly  on  the 
curve  line  AFB,  otherwise  there  would,  in 
the  one  or  the  other,  be  points  unequally  dis- 
tant from  the  centre,  which  is  contrary  to 
the  definition  of  a  circle. 


BOOK  III.  48 


PROPOSITION  II.    THEOREM. 
Every  chord  is  less  than  the  diameter. 


Let  AD  be  any  chord.  Draw  the  radii 
CA,  CD,  to  its  extremities.  We  shall  then 
have  AD<AC+CD  (Book  I.  Prop.  VII.*); 
or  AD<AB. 


Cor.     Hence  the  greatest  line  which  can  be  inscribed  in  a 
circle  is  its  diameter. 


PROPOSITION  III.    THEOREM. 

A  straight  line  cannot  meet  the  circumference  of  a  circle  in  more 
than  two  points. 

For,  if  it  could  meet  it  in  three,  those  three  points  would  be 
equally  distant  from  the  centre  ;  and  hence,  there  would  be 
three  equal  straight  lines  drawn  from  the  same  point  to  the 
same  straight  line,  which  is  impossible  (Book  I.  Prop.  XV. 
Cor.  2.). 


PROPOSITION  IV.    THEOREM. 

In  the  same  circle,  or  in  equal  circles,  equal  arcs  are  subtended  hy 
equal  chords ;  and,  conversely,  equal  chords  subtend  equal  arcs. 


Note.  When  reference  is  made  from  one  proposition  to  another,  in  the 
some  Book,  the  number  of  the  proposition  referred  to  is  alone  given ;  but  when 
the  proposition  is  found  in  a  different  Book,  the  number  of  the  Book  is  also 
given. 


44  GEOMETRY. 

If  the  radii  AC,  EO,  are 
equal,  and  also  the  arcs 
AMD,  ENG;  then  the  chord 
AD  will  be  equal  to  the 
chord  EG. 

For,  since  the  diameters 
AB,  EF,  are  equal,  the  semi- 
circle AMDB  maybe  applied 
exactly  to  the  semicircle  ENGF,  and  the  curve  line  AMDB 
will  coincide  entirely  with  the  curve  line  ENGF.  But  the 
part  AMD  is  equal  to  the  part  ENG,  by  hypothesis  ;  hence,  the 
point  D  will  fall  on  G  ;  therefore,  the  chord  AD  is  equal  to  the 
chord  EG. 

Conversely,  supposing  again  the  radii  AC,  EO,  to  be  equal, 
if  the  chord  AD  is  equal  to  the  chord  EG,  the  arcs  AMD, 
ENG  will  also  be  equal. 

For,  if  the  radii  CD,  OG,  be  drawn,  the  triangles  ACD, 
EOG,  will  have  all  their  sides  equal,  each  to  each,  namely, 
AC=:EO,  CD  =  OG,  and  AD=EG  ;  hence  the  triangles  are 
themselves  equal ;  and,  consequently,  the  angle  ACD  is  equal 
EOG  (Book  I.  Prop.  X.).  Now,  placing  the  semicircle  ADB 
on  its  equal  EGF,  since  the  angles  ACD,  EOG,  are  equal,  it  is 
plain  that  the  radius  CD  will  fall  on  the  radius  OG,  and  the 
point  D  on  the  point  G  ;  therefore  the  arc  AMD  is  equal  to  the 
arc  ENG. 


PROPOSITION  V.    THEOREM. 

In  the  same  circle^  or  in  equal  circles,  a  greater  arc  is  subtended 
by  a  greater  chord,  and  conversely,  the  greater  chord  subtends 
the  greater  arc. 

Let  the  arc  AH  be  greater  than 
the  arc  AD  ;  then  will  the  chord  AH 
be  greater  than  the  chord  AD. 

For,  draw  the  radii  CD,  CH.  The 
two  sides  AC,  CH,  of  the  triangle   .  , 
ACH  are  equal  to  the  two  AC,  CD,  -^  '' 
of  the  triangle  ACD,  and  the  angle 
ACH  is  greater  than  ACD ;  hence,  the 

third  side  AH  is  greater  than  the  third  

side  AD  (Book  I.  Prop.  IX.)  ;  there-  El 

fore  the  chord,  which  subtends  the  greater  arc,  is  the  greater. 
Conversely,  if  the  chord  AH  is  greater  than  AD,  it  will  follow, 
on  comparing  the  same  triangles,  that  the  angle  ACH    is 


BOOK  III.  45 

greater  than  ACD ;  and  hence,  that  the  arc  AH  is  greater 
than  AD. 

Scholium.  The  arcs  here  treated  of  are  each  less  than  the 
semicircumference.  If  they  were  greater,  the  reverse  pro- 
perty would  have  place  ;  for,  as  the  arcs  increase,  the  chords 
would  diminish,  and  conversely.  Thus,  the  arc  AKBD  is 
greater  than  AKBH,  and  the  chord  AD,  of  the  first,  is  less 
than  the  chord  AH  of  the  second. 


PROPOSITION  VI.    THEOREM. 

The  radius  which  is  perpendicular  to  a  chord,  bisects  the  chord, 
and  bisects  also  the  subtended  arc  of  the  chord. 

Let  AB  be  a  chord,  and  CG  the  ra- 
dius perpendicular  to  it :  then  will  AD= 
DB,  and  the  arc  AG=GB. 

For,  draw  the  radii  CA,  CB.  Then 
the  two  right  angled  triangles  ADC, 
CDB,  will  have  AC  =  CB,  and  CD  com- 
mon ;  hence,  AD  is  equal  to  DB  (Book 

I.  Prop.  xyii.). 

Again,  since  AD,  DB,  are  equal,  CG 
is  a  perpendicular  erected  from  the  mid- 
dle of  AB  ;  hence  every  point  of  this  perpendicular  must  be 
equally  distant  from  its  two  extremities  A  and  B  (Book  I.  Prop, 
XVI.).  Now,  G  is  one  of  these  points ;  therefore  AG,  BG,  are 
equal.  But  if  the  chord  AG  is  equal  to  the  chord  GB,  the  arc 
AG  will  be  equal  to  the  arc  GB  (Prop.  IV.)  ;  hence,  the  radius 
CG,  at  right  angles  to  the  chord  AB,  divides  the  arc  subtended 
by  that  chord  into  two  equal  parts  at  the  point  G. 

Scholium.  The  centre  C,  the  middle  point  D,  of  the  chord 
AB,  and  the  middle  point  G,  of  the  arc  subtended  by  this 
chord,  are  three  points  of  the  same  line  perpendicular  to  the 
chord.  But  two  points  are  sufficient  to  determine  the  position 
of  a  straight  line ;  hence  every  straight  line  which  passes  through 
two  of  the  points  just  mentioned,  will  necessarily  pass  through 
the  third,  and  be  perpendicular  to  the  chord. 

It  follows,  likewise,  that  the  perpendicular  raised  from  the 
middle  of  a  chord  passes  through  the  centre  of  the  circle,  and 
through  the  middle  of  the  arc  subtended  by  that  chord. 

For,  this  perpendicular  is  the  same  as  the  one  let  fall  from 
the  centre  on  the  same  chord,  since  both  of  them  pass  through 
the  centre  and  middle  of  the  chord. 


4ft  GEOMETRY. 


PROPOSITION  VII.    THEOREM. 

Through  three  given  points  not  in  the  same  straight  line,  one  ciV' 
cumference  rnay  always  he  made  to  pass,  and  but  one. 

Let  A,  B,  and  C,  be  the  given 
points. 

Draw  AB,  BC,  and  bisect  these 
straight  Hnes  by  the  perpendiculars 
DE,  FG :  we  say  first,  that  DE  and 
FG,  will  meet  in  some  point  O. 

For,  they  must  necessarily  cut 
each  other,  if  they  are  not  parallel. 
Now,  if  they  were  parallel,  the  line  AB,  which  is  perpendicular 
to  DE,  would  a]so  be  perpendicular  to  FG,  and  the  angle  K 
would  be  a  right  angle  (Book  I.  Prop.  XX.  Cor.  1.).  But  BK, 
the  prolongation  of  BD,  is  a  different  line  from  BF,  because  the 
three  points  A,  B,  C,  are  not  in  the  same  straight  line  ;  hence 
there  would  be  two  perpendiculars,  BF,  BK,  let  fall  from  the 
same  point  B,  on  the  same  straight  line,  which  is  impossible 
(Book  I.  Prop.  XIV.)  ;  hence  DE,  FG,  will  always  meet  in 
some  point  O. 

And  moreover,  this  point  O,  since  it  lies  in  the  perpendicular 
DE,  is  equally  distant  from  the  two  points,  A  and  B  (Book  I. 
Prop.  XVI.X ;  and  since  the  same  point  O  lies  in  the  perpen- 
dicular FG,  it  is  also  equally  distant  from  the  two  points  B  and 
C  :  hence  the  three  distances  OA,  OB,  OC,  are  equal ;  there- 
fore the  circumference  described  from  the  centre  O,  with  the 
radius  OB,  will  pass  tlirough  the  three  given  points  A,  B,  C. 

We  have  now  shown  that  one  circumference  can  always  be 
made  to  pass  through  three  given  points,  not  in  the  same 
straight  line  :  we  say  farther,  that  but  one  can  be  described 
through  them. 

For,  if  tliere  were  a  second  circumference  passing  through  the 
three  given  points  A,  B,  C,  its  centre  could  not  be  out  of  the 
line  DE,  for  then  it  would  be  unequally  distant  from  A  and  B 
(Book  I.  Prop.  XVI.);  neither  could  it  be  out  of  the  line  FG,  for 
a  like  reason  ;  therefore,  it  would  be  in  both  the  lines  DE,  FG. 
But  two  straight  lines  cannot  cut  each  other  in  more  than  one 
point ;  hence  there  is  but  one  circumference  which  can  pass 
through  three  given  points. 

Cor,  Two  circumferences  cannot  meet  in  more  than  two 
points  ;  for,  if  they  have  three  common  points,  there  would  be 
two  circumferences  passing  through  the  same  three  points ; 
which  has  been  shown  by  the  proposition  to  be  impossible. 


BOOK  III. 


47 


PROPOSITION  VIII.     THEOREM. 

Two  equal  chords  are  equally  distant  from  the  centre  ;  and  of  two 
unequal  chords,  the  less  is  at  the  greater  distance  from  the 
centre. 


First,  Suppose  the  chord  AB= 
DE.  Bisect  these  chords  by  the  per- 
pendiculars CF,  CG,  and  draw  the 
radii  CA,  CD. 

In  the  right  angled  triangles  CAF, 
DCG,  the  hypothenuses  CA,  CD,  are 
equal ;  and  the  side  AF,  the  half  of 
AB,  is  equal  to  the  side  DG,  the  half 
of  DE :  hence  the  triangles  are  equal, 
and  CF  is  equal  to  CG  (Book  I.  Prop. 
XVII.)  ;  hence,  the  two  equal  chords 
AB,  DE,  are  equally  distant  from  the  centre. 

Secondly  Let  the  chord  AH  be  greater  than  DE.  The 
arc  AKH  will  be  greater  than  DME  (Prop.  V.)  :  cut  off  from 
the  former,  a  part  ANB,  equal  to  DME  ;  draw  the  chord  AB, 
and  let  fall  CF  perpendicular  to  this  chord,  and  CI  perpendicu- 
lar to  AH.  It  is  evident  that  CF  is  greater  than  CO,  and  CO 
than  CI  (Book  I.  Prop.  XV.)  ;  therefore,  CF  is  still  greater 
than  CI.  But  CF  is  equal  to  CG,  because  the  chords  AB, 
DE,  are  equal :  hence  we  have  CG>CI;  hence  of  two  unequal 
chords,  the  less  is  the  farther  from  the  centre. 


PROPOSITION  IX.    THEOREM. 


A  straight  line  perpendicular  to  a  radius,  at  its  extremity,  is  a 
tangent  to  the  circumference. 

Let  BD  be  perpendicular  to  the   B. 
radius  C  A,  at  its  extremity  A ;  then 
will  it  be  tangent  to  the  circumfe- 
rence. 

For,  every  oblique  hne  CE,  is 
longer  than  the  perpendicular  CA 
(Book  I.  Prop.  XV.);  hence  the 
point  E  is  without  the  circle  ;  therefore,  BD  has  no  point  but 
A  common  to  it  and  the  circumference ;  consequently  BD  is  a 
tangent  (Def.  8.). 


:i8 


GEOMETRY. 


Scholium.  At  a  given  point  A,  only  one  tangent  AD  can 
be  drawn  to  the  circumference  ;  for,  if  another  could  be  drawn, 
it  would  not  be  perpendicular  to  the  radius  CA  (Book  I.  Prop, 
XIV.  Sch.) ;  hence  in  reference  to  this  new  tangent,  the  radius 
AC  would  be  an  oblique  line,  and  the  perpendicular  let  fall 
from  the  centre  upon  this  tangent  would  be  shorter  than  CA  ; 
hence  this  supposed  tangent  would  enter  the  circle,  and  be  a 
secant. 


PROPOSITION  X.    THEOREM. 


HQ  ;  in  other  words, 


Two  parallels  intercept  equal  arcs  on  the  circumfirence. 

There  may  be  three  cases. 

First.  If  the  two  parallels  are  se- 
cants, draw  the  radius  CH  perpendicu- 
lar to  the  chord  MP.  It  will,  at  the 
same  time  be  perpendicular  to  NQ 
(Book  I.Prop.XX.Cor.  1 .) ;  therefore,  the 
point  H  will  be  at  once  the  middle  of 
the  arc  MHP,  and  of  the  arc  NHQ 
(Prop.  VI.) ;  therefore,  we  shall  have 
the  arc  MH=HP,  and  the  arc  NH  = 
HQ  ;  and  therefore  MH— NH=HP- 
MN=PQ. 

Second.  When,  of  the  two  paral- 
lels AB,  DE,  one  is  a  secant,  the 
other  a  tangent,  draw  the  radius  CH 
to  the  point  of  contact  H  ;  it  will  be 
perpendicular  to  the  tangent  DE 
(Prop.  IX.),  and  also  to  its  parallel 
MP.  But,  since  CH  is  perpendicular 
to  the  chord  MP,  the  point  H  must  be 
the  middle  of  the  arc  MHP  (Prop. 
VI.) ;  therefore  the  arcs  MH,  HP,  in- 
cluded between  the  parallels  AB,  DE,  are  equal. 

Third.  If  the  two  parallels  DE,  IL,  are  tangents,  the  one 
at  H,  the  other  at  K,  draw  the  parallel  secant  AB  ;  and,  from 
what  has  just  been  shown,  we  shall  have  MH=HP,  MK=KP; 
and  hence  the  whole  arc  HMK=HPK.  It  is  farther  evident 
that  each  of  these  arcs  is  a  semicircumference. 


BOOK  IIL 


49 


PROPOSITION  XI.    THEOREM. 

If  two  circles  cut  each  other  in  two  points,  the  line  which  passes 
through  their  centres,  will  he  perpendicular  to  the  chord  which 
joins  the  points  of  intersection,  and  will  divide  it  into  two 
equal  parts. 

For,  let  the  line  AB  join  the  points  of  intersection.     It  will 
be  a  common  chord  to  the  two  circles.  Now  if  a  perpendicular 


be  erected  from  the  middle  of  this  chord,  it  will  pass  through 
each  of  the  two  centres  C  and  D  (Prop.  VI.  Sch.).  But  no 
more  than  one  straight  line  can  be  drawn  through  two  points  ; 
hence  the  straight  hne,  which  passes  through  the  centres,  will 
bisect  the  chord  at  right  angles. 


PROPOSITION  Xn.    THEOREM. 


If  the  distance  between  the  centres  of  two  circles  is  less  than  the 
sum  of  the  radii,  the  greater  radius  being  at  the  same  time 
less  than  the  sum  of  the  smaller  and  the  distance  between  the 
centres,  the  two  circumferences  will  cut  each  other. 

For,  to  make  an  intersection 
possible,  the  triangle  CAD  must 
be  possible.  Hence,  not  only 
must  we  have  CD  <  AC  +  AD, 
but  also  the  greater  radius  AD< 
AC  +  CD  (Book  I.  Prop.  VII.). 
And,  whenever  the  triangle  CAD 
can  be  constructed,  it  is  plain 
that  the  circles  described  from  the  centres  C  and  D,  will  cut 
each  other  in  A  and  B. 


5a 


GEOMETRY. 


PROPOSITION  XIII.    THEOREM. 


Ij  the  distance  between  the  centres  of  two  circles  is  equal  to  the 
sum  of  their  radiif  the  two  circles  will  touch  each  other  exter- 
nally. 


Let  C  and  D  be  the  centres  at  a 
distance  from  each  other  equal  to 
CA  +  AD. 

The  circles  will  evidently  have  the 
point  A  common,  and  they  w^ill  have 
no  other;  because,  if  they  had  two 
points  common,  the  distance  between 


their  centres  must  be  less  than  the  sum  of  their  radii. 


PROPOSITION  XIV.    THEOREM. 


If  the  distance  between  the  centres  of  two  circles  is  equal  to  the 
difference  of  their  radiif  the  two  circles  will  touch  each  other 
internally. 

Let  C  and  D  be  the  centres  at  a  dis- 
tance from  each  other  equal  to  AD — CA. 

It  is  evident,  as  before,  that  they  will 
have  the  point  A  common ;  they  can  have 
no  other;  because,  if  they  had,  the  greater 
radius  AD  must  be  less  than  the  sum  of 
the  radius  AC  and  the  distanceCD  between 
the  centres  (Prop.  XIL) ;  which  is  contraiy 
to  the  supposition. 

Cor.  Hence,  if  two  circles  touch  each  other,  either  exter- 
nally or  internally,  their  centres  and  the  point  of  contact  will 
be  in  the  same  right  line. 

Scholium.  All  circles  which  have  their  centres  on  the  right 
line  AD,  and  which  pass  through  the  point  A,  are  tangent  to 
each  other.  For,  they  have  only  the  point  A  common,  and  it 
through  the  point  A,  AE  be  drawn  perpendicular  to  AD,  the 
straight  line  AE  will  be  a  common  tangent  to  all  the  circles. 


BOOK  III.  61 


PROPOSITION  XV.    THEOREM. 


In  the  same  circle,  or  in  equal  circles,  equal  angles  having  their 
vertices  at  the  centre,  intercept  equal  arcs  on  the  circumference : 
and  conversely, -if  the  arcs  intercepted  are  equal,  the  angles 
contained  by  the  radii  will  also  he  equal. 

Let  C  and  C  be  the  centres  of  equal  circles,  and  the  angle 
ACB=DCE. 

First.  Since  the  angles  ACB, 
DCE,  are  equal,  they  may  be 
placed  upon  each  other ;  and 
since  their  sides  are  equal,  the 
point  A  will  evidently  fall  on  D, 
and  the  point  B  on  E.  But,  in 
that  case,  the  arc  AB  must  also 
fall  on  the  arc  DE  ;  for  if  the  arcs  did  not  exactly  coincide,  there 
would,  in  the  one  or  the  other,  be  points  unequally  distant  from 
the  centre  ;  which  is  impossible  :  hence  the  arc  AB  is  equal 
to  DE. 

Secondly.  If  we  suppose  AB=DE,  the  angle  ACB  will  be 
equal  to  DCE»  For,  if  these  angles  are  not  equal,  suppose 
ACB  to  be  the  greater,  and  let  ACI  be  taken  equal  to  DCE. 
From  what  has  just  been  shown,  we  shall  have  AI=DE  :  but, 
by  hypothesis,  AB  is  equal  to  DE ;  hence  AI  must  be  equal  to 
AB,  or  a  part  to  the  whole,  which  is  absurd  (Ax.  8.)  :  hence, 
the  angle  ACB  is  equal  to  DCE. 


PROPOSITION  XVI.    THEOREM. 

In  the  same  circle,  or  in  equal  circles,  if  two  angles  at  the  centre 
are  to  each  other  in  the  proportion  of  two  whole  numbers,  the 
intercepted  arcs  will  be  to  each  other  in  the  proportion  of  the 
same  numbers,  and  we  shall  have  the  angle  to  the  angle,  as  the 
corresponding  arc  to  the  corresponding  arc. 


52 


GEOMETRY. 


Suppose,  for  example,  that  the  angles  ACB,  DCE,  are  to 
each  other  as  7  is  to  4 ;  or,  which  is  the  same  thing,  suppose 
that  the  angle  M,  which  may  serve  as  a  common  measure,  is 
contained  7  times  in  the  angle  ACB,  and  4  times  in  DCE. 

C 


The  seven  partial  angles  ACw,  twCti,  nCp,  &c.,  into  which 
ACB  is  divided,  being  each  equal  to  any  of  the  four  partial 
angles  into  which  DCE  is  divided  ;  each  of  the  partial  arcs 
Am,  mn,  np,  &c.,  will  be  equal  to  each  of  the  partial  arcs  Do', 
xy,  &c.  (Prop.  XV.).  Therefore  the  whole  arc  AB  will  be  to 
the  whole  arc  DE,  as  7  is  to  4.  But  the  same  reasoning  would 
evidently  apply,  if  in  place  of  7  and  4  any  numbers  whatever 
were  employed ;  hence,  if  the  ratio  of  the  angles  ACB,  DCE, 
can  be  expressed  in  whole  numbers,  the  arcs  AB,  DE,  will  be 
to  each  other  as  the  angles  ACB,  DCE. 

Scholium,  Conversely,  if  the  arcs,  AB,  DE,  are  to  each 
other  as  two  whole  numbers,  the  angles  ACB,  DCE  will  be  to 
each  other  as  the  same  whole  numbers,  and  we  shall  have 
ACB  :  DCE  :  :  AB  :  DE.  For  the  partial  arcs.  Aw,  mw,  &c. 
and  Dx,  xy,  &c.,  being  equal,  the  partial  angles  ACm,  mCn, 
&c.  and  DCo;,  xCy,  &c.  will  also  be  equal. 


PROPOSITION  XVII.    THEOREM. 

Whatever  be  the  ratio  of  two  angles,  they  will  always  he  to  each 
other  as  the  arcs  intercepted  between  their  sides ;  the  arcs  being 
described  from  the  vertices  of  the  angles  as  centres,  with  equal 
radii. 


Let  ACB  be  the  greater  and 
ACD  the  less  angle. 

Let  the  less  angle  be  placed 
on  the  greater.  If  the  propo- 
sition is  not  true,  the  angle 
ACB  will  be  to  the  angle  ACD 
as  the  arc  AB  is  to  an  arc 
greater  or  less  than  AD.  Suppose  this  arc  to  be  greater,  and 
let  it  be  represented  by  AO  ;  we  shall  thus  have,  the  angle 
ACB  :  angle  ACD  : :  arc  AB  :  arc  AO.  Next  conceive  the  ai*c 


BOOK  III.  63 

AB  to  be  divided  into  equal  parts,  each  of  which  is  less  than 
DO  ;  there  will  be  at  least  one  point  of  division  between  D  and 
O  ;  let  I  be  that  point ;  and  draw  CI.  The  arcs  AB,  AI,  will  be 
to  each  other  as  two  whole  numbers,  and  by  the  preceding 
theorem,  we  shall  have,  the  angle  ACB  :  angle  ACI : :  arc  AB 
:  arc  AI.  Comparing  these  two  proportions  with  each  other, 
we  see  that  the  antecedents  are  the  same  :  hence,  the  conse- 
quents are  proportional  (Book  II.  Prop.  IV.) ;  and  thus  we  find 
the  angle  ACD  :  angle  ACI : :  arc  AO  :  arc  AI.  But  the  arc 
AO  is  greater  than  the  arc  AI ;  hence,  if  this  proportion  is  true, 
the  angle  ACD  must  be  greater  than  the  angle  ACI  :  on  the 
contrary,  however,  it  is  less ;  hence  the  angle  ACB  cannot  be 
to  the  angle  ACD  as  the  arc  AB  is  to  an  arc  greater  than  AD. 
By  a  process  of  reasoning  entirely  similar,  it  may  be  shown 
that  the  fourth  term  of  the  proportion  cannot  be  less  than  AD  ; 
hence  it  is  AD  itself ;  therefore  we  have 

Angle  ACB  :  angle  ACD  :  :  arc  AB  :  arc  AD. 

Cor.  Since  the  angle  at  the  centre  of  a  circle,  and  the  arc 
intercepted  by  its  sides,  have  such  a  connexion,  that  if  the  one 
be  augmented  or  diminished  in  any  ratio,  the  other  will  be 
augmented  or  diminished  in  the  same  ratio,  we  are  authorized 
to  establish  the  one  of  those  magnitudes  as  the  measure  of  the 
other ;  and  we  shall  henceforth  assume  the  arc  AB  as  the  mea- 
sure of  the  angle  ACB.  It  is  only  necessary  that,  in  the  com- 
parison of  angles  with  each  other,  the  arcs  which  serve  to 
measure  them,  be  described  with  equal  radii,  as  is  imphed  in 
all  the  foregoing  propositions. 

Scholium  1.  It  appears  most  natural  to  measure  a  quantity 
by  a  quantity  of  the  same  species ;  and  upon  this  principle  it 
would  be  convenient  to  refer  all  angles  to  the  right  angle  ; 
which,  being  made  the  unit  of  measure,  an  acute  angle  would 
be  expressed  by  some  number  between  0  and  1 ;  an  obtuse  an- 
gle by  some  number  between  1  and  2,  This  mode  of  express- 
ing angles  would  not,  however,  be  the  most  convenient  in 
practice.  It  has  been  found  more  simple  to  measure  them  by 
arcs  of  a  circle,  on  account  of  the  facility  with  which  arcs  can 
be  made  equal  to  given  arcs,  and  for  various  other  reasons.  At 
all  events,  if  the  measurement  of  angles  by  arcs  of  a  circle 
is  in  any  degree  indirect,  it  is  still  equally  easy  to  obtain  the 
direct  and  absolute  measure  by  this  method ;  since,  on 
comparing  the  arc  which  serves  as  a  measure  to  any  an- 
gle, with  the  fourth  part  of  the  circumference,  we  find  the 
ratio  of  the  given  angle  to  a  right  angle,  which  is  the  absolute 
measure. 


54 


GEOMETRY. 


Scholium  2.  All  that  has  been  demonstrated  in  the  last  three 
propositions,  concerning  the  comparison  of  angles  with  arcs, 
holds  true  equally,  if  applied  to  the  comparison  of  sectors  with 
arcs ;  for  sectors  are  not  only  equal  when  their  angles  are  so, 
but  are  in  all  respects  proportional  to  their  angles ;  hence,  two 
sectors  ACB,  ACD,  taken  in  the  same  circle,  or  in  equal  circles^ 
are  to  each  other  as  the  arcs  AB,  AD,  the  bases  of  those  sectors. 
It  is  hence  evident  that  the  arcs  of  the  circle,  which  serve  as  a 
measure  of  the  different  angles,  are  proportional  to  the  different 
sectors,  in  the  same  circle,  or  in  equal  circles. 


PROPOSITION  XVIII.     THEOREM. 

An  inscribed  angle  is  measured  by  half  the  arc  included  between 

its  sides. 


Let  BAD  be  an  inscribed  angle,  and  let 
us  first  suppose  that  the  centre  of  the  cir- 
cle lies  within  the  angle  BAD.  Draw  the 
diameter  AE,  and  the  radii  CB,  CD. 

The  angle  BCE,  being  exterior  to  the 
triangle  ABC,  is  equal  to  the  sum  of  the 
two  interior  angles  CAB,  ABC  (Book  I. 
Prop.  XXV.  Cor.  6.) :  but  the  triangle  BAC 
being  isosceles,  the  angle  CAB  is  equal  to 
ABC  ;  hence  the  angle  BCE  is  double  of  BAC.  Since  BCE  lies 
at  the  centre,  it  is  measured  by  the  arc  BE  ;  hence  BAC  will  be 
measured  by  the  half  of  BE.  For  a  like  reason,  the  angle  CAD 
will  be  measured  by  the  half  of  ED ;  hence  BAC  +  CAD,  or  BAD 
will  be  measured  l?y  half  of  BE  +  ED,  or  of  BED. 

Suppose,  in  the  second  place,  that  the 
centre  C  lies  without  the  angle  BAD.  Then 
drawing  the  diameter  AE,  the  angle  BAE 
will  be  measured  by  the  half  of  BE  ;  the 
angle  DAE  by  the  half  of  DE  :  hence  their 
difi'erence  BAD  will  be  measured  by  the 
half  of  BE  minus  the  half  of  ED,  or  by  the 
half  of  BD. 

Hence  every  inscribed  angle  is  measured 
by  half  of  the  arc  included  between  its  sides* 


BOOK  III. 


55 


Cor.  1.  All  the  angles  BAG,  BDC, 
BEC,  inscribed  in  the  same  segment  are 
equal ;  because  they  are  all  measured  by 
the  half  of  the  same  arc  BOC. 


Cor.  2.  Every  angle  BAD,  inscribed  in  a 
semicircle  is  a  right  angle  ;  because  it  is  mea- 
sured by  half  the  semicircumference  BOD, 
that  is,  by  the  fourth  part  of  the  whole  cir- 
cumference. 


Cor.  3.  Every  angle  BAG,  inscribed  in  a 
segment  greater  than  a  semicircle,  is  an  acute 
angle  ;  for  it  is  measured  by  half  of  the  arc 
BOG,  less  than  a  semicircumference. 

And    every  angle  BOG,  inscribed   in  a 
segment  less  than  a  semicircle,  is  an  obtuse 
angle  ;  for  it  is  measured  by  half  of  the  arc  B 
BAG,  greater  than  a  semicircumference. 


Cor.  4.  The  opposite  angles  A  and  G,  of 
an  inscribed  quadrilateral  ABGD,  are  to- 
gether equal  to  two  right  angles  :  for  the  an- 
gle BAD  is  measured  by  half  the  arc  BGD, 
the  angle  BGD  is  measured  by  half  the  arc 
BAD  ;  hence  the  two  angles  BAD,  BGD,  ta- 
ken together,  are  measured  by  the  half  of  the 
circumference  ;  hence  their  sum  is  equal  to  two  right  angles. 


PROPOSITION  XIX.    THEOREM. 


The  angle  formed  hy  two  chords,  which  intersect  each  other,  is 
measured  by  half  the  sum  of  the  arcs  included  between  its  sides. 


56 


GEOMETRY. 


Let  AB,  CD,  be  two  chords  intersecting 
each  other  at  E :  then  will  the  angle 
AEC,  or  DEB,  be  measured  by  half  of 
AC  +  DB. 

Draw  AF  parallel  to  DC :  then  will 
the  arc  DF  be  equal  to  A(^  (Prop.  X.)  ; 
and  the  angle  FAB  equal  to  the  angle 
DEB  (Book  I.  Prop.  XX.  Cor.  3.).  But 
the  angle  FAB  is  measured  by  half  the 
arc  FDB  (Prop.  XVIII.);  therefore,  DEB 
is  measured  by  half  of  FDB ;  that  is,  by  half  of  DB  +  DF,  or 
half  of  DB  +  AC.  In  the  same  manner  it  might  be  proved  that 
the  angle  AED  is  measured  by  half  of  AFD+BC. 


PROPOSITION  XX.    THEOREM. 


The  angle  formed  hy  two  secants,  is  measured  by  half  the  diffe- 
rence of  the  arcs  included  between  its  sides. 


Let  AB,  AC,  be  two  secants  :  then 
will  the  angle  BAC  be  measured  by 
half  the  difference  of  the  arcs  BEC 
and  DF. 

Draw  DE  parallel  to  AC  :  then  will 
the  arc  EC  be  equal  to  DF,  and  the 
angle  BDE  equal  to  the  angle  BAC. 
But  BDE  is  measured  by  half  the  arc 
BE ;  hence,  BAC  is  also  measured  by 
half  the  arc  BE  ;  that  is,  by  half  the 
difference  of  BEC  and  EC,  or  half  the 
difference  of  BEC  and  DF. 


PROPOSITION  XXI.    THEOREM. 


The  angle  formed  by  a  tangent  and  a  chord,  is  measured  by  half 
of  the  arc  included  between  its  sides. 


BOOK  III. 


57 


Let  BE  be  the  tangent,  and  AC  the  chord. 

From  A,  the  point  of  contact,  draw  the 
diameter  AD.  The  angle  BAD  is  a  right 
angle  (Prop.  IX.),  and  is  measured  by 
half  the  semi  circumference  AMD ;  the 
angle  DAC  is  measured  by  the  half  of 
DC:  hence,  BAD  +  DAC,  or  BAC,  is 
measured  by  the  half  of  AMD  plus  the 
half  of  DC,  or  by  half  the  whole  arc 
AMDC. 

It  might  be  shown,  by  taking  the  difference  between  the  an- 
gles DAE,  DAC,  that  the  angle  CAE  is  measured  by  half  the 
arc  AC,  included  between  its  sides. 


'.m9^9* 


PROBLEMS  RELATING  TO  THE  FIRST  AND  THIRD  BOOKS. 


PROBLEM  I. 


To  divide  a  given  straight  line  into  two  equal  parts. 


X> 


4^ 


Let  AB  be  the  given  straight  line.     " 

From  the  points  A  and  B  as  centres,  with 
a  radius  greater  than  the  half  of  AB,  describe 
two  arcs  cutting  each  other  in  D  ;  the  point 
D  will  be  equally  distant  from  A  and  B.  Find, 
in  like  manner,  above  or  beneath  the  line  AB,  -j 
a  second  point  E,  equally  distant  from  the  " 
points  A  and  B ;  through  the  two  points  D 
and  E,  draw  the  line  DE :  it  will  bisect  the 
hne  AB  in  C. 

For,  the  two  points  D  and  E,  being  each  equally  distant  from 
the  extremities  A  and  B,  must  both  lie  in  the  perpendicular 
raised  from  the  middle  of  AB  (Book  I.  Prop.  XVI.  Cor.).  But 
only  one  straight  line  can  pass  through  two  given  points  ;  hence 
the  line  DE  must  itself  be  that  perpendicular,  which  divides 
AB  into  two  equal  parts  at  the  point  C. 


><^ 


8 


58 


GEOMETRY. 


^       * 


PROBLEM  II. 

At  a  given  point,  in  a  given  straight  line,  to  erect  a  perpendicu- 
lar to  this  line. 


>^ 


Let  A  be  the  given  point,  and  BC  the 
given  Hne. 

Take  the  points  B  and  C  at  equal  dis- 
tances from  A  ;  then  from  the  points  B  and 
C  as  centres,  with  a  radius  greater  than  — t— 
BA,  describe  two  arcs  intersecting  each 
other  in  D ;  draw  AD  :   it  will  be  the  perpendicular  required. 

For,  the  point  D,  being  equally  distant  from  B  and  C,  must 
be  in  the  perpendicular  raised  from  the  middle  of  BC  (Book  L 
Prop.  XVI.) ;  and  since  two  points  determine  a  line,  AD  is  that 
perpendicular. 

Scholium.  The  same  construction  serves  for  making  a  right 
angle  BAD,  at  a  given  point  A,  on  a  given  straight  line  BC. 


PROBLEM  III. 

From  a  given  point,  without  a  straight  line,  to  let  fall  a  perpen- 
dicular on  this  line. 


Let  A  be  the  point,  and  BD  the  straight 
line. 

From  the  point  A  as  a  centre,  and  with 
a  radius  sufficiently  great,  describe  an 
arc  cutting  the  line  BD  in  the  two  points 
B  and  D  ;  then  mark  a  point  E,  equally 
distant  from  the  points  B  and  D,  and 
draw  AE  :  it  will  be  the  perpendicular  required. 

For,  the  two  points  A  and  E  are  each  equally  distant  from 
the  points  B  and  D ;  hence  the  line  AE  is  a  perpendicular 
passing  through  the  middle  of  BD  (Book  L  Prop.  XVL  Cor.). 


PROBLEM   IV. 


At  a  point  in  a  given  line,  to  make  an  angle  equal  to  a  given 

angle. 


BOOK  III.  69 

Let  A  be  the  given  point,  AB  the  given  line,  and  IKL  the 
given  angle. 

From  the  vertex  K,  as  a  cen-  t  o<^ 

tre,  with  any  radius,  describe  the  ^^^  ^^^\ 

arc  IL,  terminating  in  the  two        ^y^  \      ^y^    \ 

sides  of  the  angle.     From  the    E-^^- XA  B 

point  A  as  a  centre,  with  a  dis- 
tance AB,  equal  to  KI,  describe  the  indefinite  arc  BO  ;  then 
take  a  radius  equal  to  the  chord  LI,  with  which,  from  the  point 
B  as  a  centre,  describe  an  arc  cutting  the  indefinite  arc  BO,  in 
D ;  draw  AD ;  and  the  angle  DAB  will  be  equal  to  the  given 
angle  K. 

For,  the  two  arcs  BD,  LI,  have  equal  radii,  and  equal  chords ; 
hence  they  are  equal  (Prop.  IV.) ;  therefore  the  angles  BAD, 
IKL,  measured  by  them,  are  equal. 


PROBLEM  V. 
To  divide  a  given  arc,  or  a  given  angle,  into  two  equal  parts. 

First.  Let  it  be  required  to  divide  the 
arc  AEB  into  two  equal  parts.  From  the 
points  A  and  B,  as  centres,  with  the  same 
radius,  describe  two  arcs  cutting  each  other 
in  D  ;  through  the  point  D  and  the  centre 
C,  draw  CD :  it  will  bisect  the  arc  AB  in 
the  point  E. 

For,  the  two  points  C  and  D  are  each 
equally  distant  from  the  extremities  A  and 
B  of  the  chord  AB  ;  hence  the  line  CD  bi- 
sects the  chord  at  right  angles   (Book  I.  Prop.  XVI.  Cor.)  ; 
hence,  it  bisects  the  arc    AB  in  the    point  E   (Prop.  VI.). 

Secondly.  Let  it  be  required  to  divide  the  angle  ACB  into 
two  equal  parts.  We  begin  by  describing,  from  the  vertex  C 
as  a  centre,  the  arc  AEB ;  which  is  then  bisected  as  above.  It 
is  plain  that  the  line  CD  will  divide  the  angle  ACB  into  two 
equal  parts. 

Scholium.  By  the  same  construction,  each  of  the  halves 
AE,  EB,  may  be  divided  into  two  equal  parts  ;  and  thus,  by 
successive  subdivisions,  a  given  angle,  or  a  given  arc  may 
be  divided  into  four  equal  parts,  into  eight,  into  sixteen, 
and  so  on. 


60 


GEOMETRY. 


PROBLEM  VI. 


Through  a  given  point,  to  draw  a  parallel  to  a  given  straight 

line. 


Let  A  be  the  given  point,  and  BC 
the  given  line.  B 

From  the  point  A  as  a  centre,  with 
a  radius  greater  than  the  shortest  dis- 
tance from  A  to  BC,  describe  the  in- 
definite arc  EO ;  from  the  point  E  as 
a  centre,  with  the  same  radius,  describe  the  arc  AF  ;  make 
ED=AF,  and  draw  AD :  this  will  be  the  parallel  required. 

For,  drawing  AE,  the  alternate  angles  AEF,  EAD,  are  evi- 
dently equal ;  therefore,  the  lines  AD,  EF,  are  parallel  (Book  I. 
Prop.  XIX.  Cor.  1.). 


E 


Dl 

o 


PROBLEM  VII. 


Two  angles  of  a  triangle  being  given,  to  find  the  third. 

Draw  the  indefinite  line  DEF; 
at  any  point  as  E,  make  the  an- 
gle DEC  equal  to  one  of  the 
given  angles,  and  the  angle 
CEH  equal  to  the  other :  the 
remaining  angle  HEF  will  be 
the  third  angle  required ;  be- 
cause those  three  angles  are 

together  equal  to  two  right  angles   (Book  I.  Prop.  I.  and 
XXV). 

problem'  VIII. 


Two  sides  of  a  triangle,  and  the  angle  which  they  contain,  being 
given,  to  describe  the  triangle. 

Let  the  lines  B  and  C  be  equal  to 
the  given  sides,  and  A  the  given  an- 
gle. 

Having  drawn  the  indefinite  line 
DE,  at  the  point  D,  make  the  angle 
EDF  equal  to  the  given  angle  A  ; 
then  take  DG=B,  DH=C,  and  draw  GH ;  DGH  will  be  the 
triangle  required  (Book  L  Prop.  V.). 


BOOK  III. 


61 


PROBLEM  IX. 


A  side  and  two  angles  of  a  triangle  being  given,  to  describe  the 

triangle. 

The  two  angles  will  either  be  both  ad- 
jacent to  the  given  side,  or  the  one  adja- 
cent, and  the  other  opposite :  in  the  lat- 
ter case,  find  the  third  angle  (Prob. 
VII.) ;  and  the  two  adjacent  angles  will 
thus  be  known  :  draw  the  straight  line 
DE  equal  to  the  given  side  :  at  the  point  D,  make  an  angle 
EDF  equal  to  one  of  the  adjacent  angles,  and  at  E,  an  angle 
DEG  equal  to  the  other ;  the  two  lines  DF,  EG,  will  cut  each 
other  in  H  ;  and  DEH  will  be  the  triangle  required  (Book  I. 
Prop.  VI.). 


PROBLEM  X. 


The  three  sides  of  a  triangle  being  given,  to  describe  the  triangle^ 

Let  A,  B,  and  C,  be  the  sides. 

Draw  DE  equal  to  the  side  A ; 
from  the  point  E  as  a  centre,  with 
a  radius  equal  to  the  second  side  B, 
describe  an  arc  ;  from  D  as  a  cen- 
tre, with  a  radius  equal  to  the  third 
side  C,  describe  another  arc  inter- 
secting the  former  in  F ;  draw  DF, 
EF ;  and  DEF  will  be  the  triangle 
required  (Book  I.  Prop.  X.). 

Scholium.  If  one  of  the  sides  were  greater  than  the  sum  of 
the  other  two,  the  arcs  would  not  intersect  each  other  :  but  the 
solution  will  always  be  possible,  when  the  sum  of  two  sides,  any 
how  taken,  is  greater  than  the  third. 


62 


GEOMETRY. 


PROBLEM  XI. 


Two  sides  of  a  triangle,  and  the  angle  opposite  one  of  them,  being 
given,  to  describe  the  triangle. 

Let  A  and  B  be  the  given  sides,  and  C  the  given  angle. 
There  are  two  cases. 

First.  When  the  angle  C  is  a  right 
angle,  or  when  it  is  obtuse,  make 
the  angle  EDF=C;  take  DE=A  ; 
from  the  point  E  as  a  centre, 
with  a  radius  equal  to  the  given 
side  B,  describe  an  arc  cutting  DF 
in  F;  draw  EF :  then  DEF  will  be 
the  triangle  required. 

In  this  first  case,  the  side  B  must 
be  greater  than  A  ;  for  the  angle  C, 
being  a  right  angle,  or  an  obtuse  an- 
gle, is  the  greatest  angle  of  the  tri- 
angle, and  the  side  opposite  to  it  must,  therefore,  also  be  the 
greatest  (Book  I.  Prop.  XIII.). 

Ai \ 


Secondly.  If  the  angle  C  is 
acute,  and  B  greater  than  A,  the 
same  construction  will  again  ap- 
ply, and  DEF  will  be  the  triangle 
required. 


But  if  the  angle  C  is  acute,  and 
the  side  B  less  than  A,  then  the 
arc  described  from  the  centre  E, 
with  the  radius  EF=B,  will  cut 
the  side  DF  in  two  points  F  and 
G,  lying  on  the  same  side  of  D : 
hence  there  will  be  two  triangles 
DEF,  DEG,  either  of  which  wiU 
satisfy  the  conditions  of  the  pro- 
blem. 


Scholium.  If  the  arc  described  with  E  as  a  centre,  should 
be  tangent  to  the  line  DG,  the  triangle  would  be  right  angled, 
and  there  would  be  but  one  solution.  The  problem  would  be 
impossible  in  all  cases,  if  the  side  B  were  less  than  the  perpen- 
dicular let  fall  from  E  on  the  line  DF. 


4- 


BOOK  III. 


PROBLEM  XII. 


The  adjacent  sides  of  a  parallelogram^  with  the  angle  which  they 
contain^  being  given^  to  describe  the  parqllelogram. 

Let  A  and  B  be  the  given  sides,  and  C  the  given  angle. 

Draw  the  line  DE=A;  at  the 
point  D,  make  the  angle  EDF= 
C  ;  take  DF=B  ;  describe  two 
arcs,  the  one  from  F  as  a  cen- 
tre, with  a  radius  FG=DE,  the 
other  from  E  as  a  centre,  with 
a  radius  EG=DF;  to  the  point 
G,  where  these  arcs  intersect 
each  other,  draw  FG,  EG ; 
DEGF  will  be  the  parallelogram  required. 

For,  the  opposite  sides  are  equal,  by  construction  ;  hence  the 
figure  is  a  parallelogram  (Book  I.  Prop.  XXIX.)  :  and  it  is 
formed  with  the  given  sides  and  the  given  angle. 

Cor.  If  the  given  angle  is  a  right  angle,  the  figure  will  be 
a  rectangle  ;  if,  in  addition  to  this,  the  sides  are  equal,  it  will 
be  a  square.  < 


PROBLEM  Xin. 

To  find  the  centre  of  a  given  circle  or  arc. 

Take  three  points,  A,  B,  C,  any 
where  in  the  circumference,  or  the 
arc;  draw  AB,BC,  or  suppose  them 
to  be  drawn ;  bisect  those  two  lines 
by  the  perpendiculars  DE,  FG : 
the  point  O,  where  these  perpen- 
diculars meet,  will  be  the  centre 
sought  (Prop.  VI.  Sch.). 

Scholium.  The  same  construc- 
tion serves  for  making  a  circum- 
ference pass  through  three  given  points  A,  B,  C  ;  and  also  for 
describing  a  circumference,  in  which,  a  given  triangle  ABC 
shall  be  inscribed. 


^ir- 


^^■. 


64 


GEOMETRY. 


PROBLEM  XIV. 
Through  a  given  pointy  to  draw  a  tangent  to  a  given  circle. 


If  the  given  point  A  lies  in  the  circum-  ^ 
ference,  draw  the  radius  CA,  and  erect 
AD  perpendicular  to  it :  AD  will  be  the 
tangent  required  (Prop.  IX.).  ^     . 


If  the  point  A  lies  without  the  circle, 
join  A  and  the  centre,  by  the  straight 
line  CA :  bisect  CA  in  O  ;  from  O  as  a 
centre,  with  the  radius  OC,  describe  a 
circumference  intersecting  the  given  cir- 
cumference in  B ;  draw  AB  :  this  will  be 
the  tangent  required. 

For,  drawing  CB,  the  angle  CBA  be- 
ing inscribed  in  a  semicircle  is  a  right 
angle  (Prop.  XVIII.  Cor.  2.) ;  therefore 
AB  is  a  perpendicular  at  the  extremity 
of  the  radius  CB  ;  therefore  it  is  a  tan- 
gent. 

Scholium.  When  the  point  A  lies  without  the  circle,  there 
will  evidently  be  always  two  equal  tangents  AB,  AD,  passing 
through  the  point  A :  they  are  equal,  because  the  right  angled 
triangles  CBA,  CDA,  have  the  hypothenuse  CA  common,  and 
the  side  CB  =  CD;  hence  they  are  equal  (Book  I.  Prop.  XVII.); 
hence  AD  is  equal  to  AB,  and  also  the  angle  CAD  to  CAB. 
And  as  there  can  be  but  one  line  bisecting  the  angle  BAC,  it 
follows,  that  the  line  which  bisects  the  angle  formed  by  two 
tangents^  must  pass  through  the  centre  of  the  circle. 


PROBLEM  XV. 
To  inscribe  a  circle  in  a  given  triangle. 


Let  ABC  be  the  given  triangle. 

Bisect  the  angles  A  and  B,  by 
the  lines  AO  and  BO,  meeting  in 
the  point  O  ;  from  the  point  O, 
let  fall  the  perpendiculars  OD, 
OE,  OF,  on  the  three  sides  of  the 
triangle:  these  perpendiculars  will 
all  be  equal.     For,  by  construe- 


BOOK  III.        ^  66 

tion,  we  have  the  angle  DAO=OAF,  the  right  angle  ADO  = 
AFO  ;  hence  the  third  angle  AOD  is  equal  to  the  third  AOF 
(Book  I.  Prop.  XXV.  Cor.  2.).  Moreover,  the  side  AO  is  com- 
mon to  the  two  triangles  AOD,  AOF ;  and  the  angles  adjacent 
to  the  equal  side  are  equal :  hence  the  triangles  themselves  are 
equal  (Book  I.  Prop.  VI.) ;  and  DO  is  equal  to  OF.  In  the  same 
manner  it  may  be  shown  that  the  two  triangles  BOD,  BOE, 
are  equal ;  therefore  OD  is  equal  to  OE  ;  therefore  the  three 
perpendiculars  OD,  OE,  OF,  are  all  equal. 

Now,  if  from  the  point  O  as  a  centre,  with  the  radius  OD, 
a  circle  be  described,  this  circle  will  evidently  be  inscribed  in 
the  triangle  ABC  ;  for  the  side  AB,  being  perpendicular  to  the 
radius  at  its  extremity,  is  a  tangent ;  and  the  same  thing  is  true 
of  the  sides  BC,  AC. 

'  -  Scholium.  The  three  lines  which  bisect  the  angles  of  a  tri- 
angle meet  in  the  same  point. 


PROBLEM  XVI. 

On  a  given  straight  line  to  describe  a  segment  that  shall  contain 
a  given  cCngle ;  that  is  to  say,  a  segment  such,  that  all  the  an- 
gles inscribed  in  it,  shall  be  equal  to  the  given  angle. 

Let  AB  be  the  given  straight  Hne,  and  C  the  given  angle. 


Produce  AB  towards  D ;  at  the  point  B,  make  the  angle 
DBE  =  C  ;  draw  BO  perpendicular  to  BE,  and  GO  perpen- 
dicular to  AB,  through  the  middle  point  G ;  and  from  the  point 
O,  where  these  perpendiculars  meet,  as  a  centre,  with  a  dis- 
tance OB,  describe  a  circle :  the  required  segment  will  be 
AMB. 

For,  since  BF  is  a  perpendicular  at  the  extremity  of  the 
radius  OB,  it  is  a  tangent,  and  the  angle  ABF  is  measured  by 
half  the  arc  AKB  (Prop.  XXL).  Also,  the  angle  AMB,  being 
an  inscribed  angle,  is  measured  by  half  the  arc  AKB  :  hence 
we  have  AMB=ABF=EBD  =  C ;  hence  all  the  angles  in- 
scribed in  the  segment  AMB  are  equal  to  the  given  angle  C. 

F*9 


66  GEOMETRY. 

Scholium.  If  the  given  angle  were  a  right  angle,  the  required 
segment  would  be  a  semicircle^  described  on  AB  as  a  diameter. 


PROBLEM  XVII. 

To  find  the  numerical  ratio  of  two  given  straight  lines,  these  lines 
being  supposed  to  have  a  common  measure. 

Let  Afi  and  CD  be  the  given  lines.  A  ,  C 

From  the  greater  AB  cut  off  a  part  equal  to  the  less 
CD,  as  many  times  as  possible  ;  for  example,  twice, 
with  the  remainder  BE. 

From  the  line  CD,  cut  off  a  part  equal  to  the  re- 
mainder BE,  as  many  times  as  possible  ;  once,  for  ex- 
ample, with  the  remainder  DP. 

From  the  first  remainder  BE,  cut  off  a  part  equal  to 
the  second  DF,  as  many  times  as  possible  ;  once,  for 
example,  with  the  remainder  BG. 

From  the  second  remainder  DF,  cut  off  a  part  equal    J.0. 
to  BG  the  third,  as  many  times  as  possible. 

Continue  this  process,  till  a  remainder  occurs,  which    "° 
is  contained  exactly  a  certain  number  of  times  in  the  preced- 
ing one. 

Then  this  last  remainder  will  be  the  common  measure  of  the 
proposed  lines ;  and  regarding  it  as  unity,  we  shall  easily  find 
the  values  of  the  preceding  remainders ;  and  at  last,  those  of 
the  two  proposed  lines,  and  hence  their  ratio  in  numbers. 

Suppose,  for  instance,  we  find  GB  to  be  contained  exactly 
twice  in  FD  ;  BG  will  be  the  common  measure  of  the  two  pro- 
posed hues.  Put  BG=1 ;  we  shall  have  FD=2  :  but  EB  con- 
tains FD  once,  plus  GB ;  therefore  we  have  EB=3  ;  CD  con- 
tains EB  once,  plus  FI) ;  therefore  we  have  CD=5  :  and, 
lastly,  AB  contains  CD  twice,  plus  EB  ;  therefore  we  have 
AB  =  13  ;  hence  the  ratio  of  the  lines  is  that  of  13  to  5.  If  the 
line  CD  were  taken  for  unity,  the  line  AB  would  be  '/  ;  if  AB 
were  taken  for  unity,  CD  would  be  /g . 


--B 


Scholium.  The  method  just  explained  is  the  same  as  that 
employed  in  arithmetic  to  find  the  common  divisor  of  two  num- 
bers :  it  has  no  need,  therefore,  of  any  other  demonstration. 

How  far  soever  the  operation  be  continued,  it  is  possible 
that  no  remainder  may  ever  be  found,  which  shall  be  contained 
an  exact  number  of  times  in  the  preceding  one.  When  this 
happens,  the  two  lines  have  no  common  measure,  and  are  said 
to  be  incommensurable.  An  instance  of  this  will  be  seen  after- 


BOOK  III.  67 

wards,  in  the  ratio  of  the  diagonal  to  the  side  of  the  square. 
In  those  cases,  therefore,  the  exact  ratio  in  numbers  cannot  be 
found  ;  but,  by  neglecting  the  last  remainder,  an  approximate 
ratio  will  be  obtained,  more  or  less  correct,  according  as  the 
operation  has  been  continued  a  greater  or  less  number  of  times. 


PROBLEM  XVIII. 

Two  angles  being  given,  to  find  their  common  measure,  if  they 
have  one,  and  by  means  of  it,  their  ratio  in  numbers. 

Let  A  and  B  be  the  given  an- 

With  equal  radii  describe  the 
arcs  CD,  EF,  to  serve  as  mea- 
sures for  the  angles :  proceed 
afterwards  in  the  comparison  of 
the  arcs  CD,  EF,  as  in  the  last 

problem,  since  an  arc  may  be  cut  off  from  an  arc  of  the  same 
radius,  as  a  straight  line  from  a  straight  line.  We  shall  thus 
arrive  at  the  common  measure  of  the  arcs  CD,  EF,  if  they  have 
one,  and  thereby  at  their  ratio  in  numbers.  This  ratio  will  be 
the  same  as  that  of  the  given  angles  (Prop.  XVII.) ;  and  if  DO 
is  the  common  measure  of  the  arcs,  DAO  will  be  that  of  the 
angles. 

Scholium.  According  to  this  method,  the  absolute  value  of 
an  angle  may  be  found  by  comparing  the  arc  which  measures 
it  to  the  whole  circumference.  If  the  arc  CD,  for  example,  is 
to  the  circumference,  as  3  is  to  25,  the  angle  A  will  be  2^5  of  four 
right  angles,  or  ^|  of  one  right  angle. 

It  may  also  happen,  that  the  arcs  compared  have  no  com- 
mon measure  ;  in  which  case,  the  numerical  ratios  of  the  angles 
will  only  be  found  approximatively  with  more  or  less  correct- 
ness, according  as  the  operation  has  been  continued  a  greater 
or  less  number  of  times. 


68 


GEOMETRY. 


BOOK  IV. 


OE  THE  PROPORTIONS  OF  FIGURES,  AND  THE  MEASUREMENT 
OF  AREAS. 


Definitions. 


1.  Similar  figures  are  those  which  have  the  angles  of  the  one 
equal  to  the  angles  of  the  other,  each  to  each,  and  the  sides 
about  the  equal  angles  proportional. 

2.  Any  two  sid(ite,  or  any  two  angles,  which  have  like  po- 
sitions in  two  similar  figures,  are  called  homologous  sides  or 
angles. 

3.  In  two  different  circles,  similar  arcSf  sectors^  or  segments f 
are  those  which  correspond  to  equal  angles  at  the  centre. 

Thus,  if  the  angles  A  and  O  are  equal, 
the  arc  BC  will  be  similar  to  DE,  the 
sector  BAG  to  the  sector  DOE,  and  the 
segment  whose  chord  is  BC,  to  the  seg- 
ment whose  chord  is  DE. 

4.  The  base  of  any  rectilineal  figure,  is  the  side  on  which 
the  figure  is  supposed  to  stand. 

5.  The  altitude  of  a  triangle  is  the  per- 
pendicular let  fall  from  the  vertex  of  an 
angle  on  the  opposite  side,  taken  as  a 
base.  Thus,  AD  is  the  altitude  of  the 
triangle  BAG 


D  E 


P   B 


DEC 


6.  The  altitude  of  a  parallelogram  is  the 
perpendicular  which  measures  the  distance 
between  two  opposite  sides  taken  as  bases. 
Thus,  EF  is  the  altitude  of  the  parallelo-  j 
gram  DB. 

7.  The  altitude  of  a  trapezoid  is  the  per- 
pendicular drawn  between  its  two  parallel 
sides.  Thus,  EF  is  the  altitude  of  the  trape- 
zoid DB. 

jtV.         JtJ'  u 

8.  The  area  and  surface  of  a  figure,  are  terms  very  nearly 
synonymous.  The  area  designates  more  particularly  the  super- 
ficial content  of  the  figure.     The  area  is  expressed  numeri- 


L        P         B 


BOOK  IV.  69 

cally  by  the  number  of  times  which  the  figure  contains  some 
other  area,  that  is  assumed  for  its  measuring  unit. 

9.  Figures  have  equal  areas,  when  they  contain  the  same 
measuring  unit  an  equal  number  of  times. 

10.  Figures  which  have  equal  areas  are  called  equivalent. 
The  term  equal,  when  applied  to  figures,  designates  those  which 
are  equal  in  every  respect,  and  which  being  applied  to  each 
other  will  coincide  in  all  their  parts  (Ax.  13.)  :  the  term  equi- 
valent  implies  an  equality  in  one  respect  only ;  namely,  an 
equality  between  the  measures  of  figures. 

We  may  here  premise,  that  several  of  the  demonstrations 
are  grounded  on  some  of  the  simpler  operations  of  algebra, 
which  are  themselves  dependent  on  admitted  axioms.  Thus, 
if  we  have  A=B  +  C,  and  if  each  member  is  multiplied  by  the 
same  quantity  M,  we  may  infer  that  AxM=BxM  +  CxM; 
in  like  manner,  if  we  have,  A=B-|-C,  and  D=E — C,  and  if  the 
equal  quantities  are  added  together,  then  expunging  the  +  C 
and  — C,  which  destroy  each  other,  we  infer  that  A  +  D=B-{- 
E,  and  so  of  others.  AH  this  is  evident  enough  of  itself;  but 
in  cases  of  difficulty,  it  will  be  useful  to  consult  some  agebrai- 
cal  treatise,  and  thus  to  combine  the  study  of  the  two  sciences. 


PROPOSITION  I.     THEOREM. 

Parallelograms  which  have  equal  bases  and  equal  altitudes,  are 

equivalent. 

Let  AB  be  the  common  base  of-j)  CT  EDPCE 
the  two  parallelograms  ABCD,  V  y  /  Vl  T? 
ABEF:  and  since  they  are  sup-     \     /\    /       \  /       \  / 

posed  to  have  the  same  altitude,         v_V  V V 

their  upper  bases  DC,  FE,  will  be        ^       ^  A        B 

both  situated  in  one  straight  line  parallel  to  AB. 

Now,  from  the  nature  of  parallelograms,  we  have  AD=BC, 
andAF=BE;  for  the  same  reason,  we  have  DC=AB,  and 
FE=AB;  hence  DC=FE:  hence,  if  DC  and  FE  be  taken 
away  from  the  same  line  DE,  the  remainders  CE  and  DF  will 
be  equal :  hence  it  follows  that  the  triangles  DAF,  CBE,  are 
mutually  eqilateral,  and  consequently  equal  (Book  I.  Prop.  X.). 

But  if  from  the  quadrilateral  ABED,  we  take  away  the  tri- 
angle ADF,  there  will  remain  the  parallelogram  ABEF ;  and 
if  from  the  same  quadrilateral  ABED,  we  take  away  the  equal 
triangle  CBE,  there  will  remain  the  parallelogram  ABCD. 


70 


GEOMETRY. 


Hence  these  two  parallelograms  ABCD,  ABEF,  which  have 
the  same  base  and  altitude,  are  equivalent. 

Cor.  Every  parallelogram  is  equivalent  to  the  rectangle      I 
which  has  the  same  base  and  the  same  altitude. 


PROPOSITION  II.    THEOREM. 

Every  triangle  is  half  the  parallelogram  which  has  the  same  hose 
and  the  same  altitude. 

Let  ABCD  be  a  parallelo- 
gram, and  ABE  a  triangle, 
having  the  same  base  AB, 
and  the  same  altitude  :  then 
will  the  triangle  be  half  the 
parallelogram.  •^  -^ 

For,  since  the  triangle  and  the  parallelogram  have  the  same 
altitude,  the  vertex  E  of  the  triangle,  will  be  in  the  line  EC,  par- 
allel to  the  base  AB,  Produce  BA,  and  from  E  draw  EF 
parallel  to  AD.  The  triangle  FBE  is  half  the  parallelogram 
FC,  and  the  triangle  FAE  half  the  parallelogram  FD  (Book  I. 
Prop.  XXVIII.  Cor.). 

Now,  if  from  the  parallelogram  FC,  there  be  taken  the  par- 
allelogram FD,  there  will  remain  the  parallelogram  AC  :  and 
if  from  the  triangle  FBE,  which  is  half  the  first  parallelogram, 
there  be  taken  the  triangle  FAE,  half  the  second,  there  will  re- 
main the  triangle  ABE,  equal  to  half  the  parallelogram  AC. 

Cor  1.  Hence  a  triangle  ABE  is  half  of  the  rectangle  ABGH, 
which  has  the  same  base  AB,  and  the  same  altitude  AH  :  for 
the  rectangle  ABGH  is  equivalent  to  the  parallelogram  ABCD 
(Prop.  I.  Cor.). 


Cor.  2.  All  triangles,  which  have  equal  bases  and  altitudes, 
are  equivalent,  being  halves  of  equivalent  parallelograms. 


PROPOSITION  III.    THEOREM. 

Two  rectangles  having  the  same  altitude^  are  to  each  other  as  their 

bases. 


m. 


BOOK  IV. 


71 


D 

!&             C 

1 

A 

E 

B 

Let  ABCD,  AEFD,  be  two  rectan- 
gles having  the  common  altitude  AD : 
they  are  to  each  other  as  their  bases 
AB,  AE. 

Suppose,  first,  that  the  bases  are 
commensurable,  and  are  to  each  other, 

for  example,  as  the  numbers  7  and  4.  If  AB  be  divided  into  7 
equal  parts,  AE  will  contain  4  of  those  parts :  at  each  point  of 
division  erect  a  perpendicular  to  the  base  ;  seven  partial  rect- 
angles will  thus  be  formed,  all  equal  to  each  other,  because  all 
have  the  same  base  and  altitude.  The  rectangle  ABCD  will 
contain  seven  partial  rectangles,  while  AEFD  will  contain  four; 
hence  the  rectangle  ABCD  is  to  AEFD  as  7  is  to  4,  or  as  AB 
is  to  AE.  The  same  reasoning  may  be  applied  to  any  other 
ratio  equally  with  that  of  7  to  4  :  hence,  whatever  be  that  ratio, 
if  its  terms  be  commensurable,  we  shall  have 

ABCD  :  AEFD  :  :  AB  :  AE. 

Suppose,  in  the  second  place,  that  the  bases 
AB,  AE,  are  incommensurable  :  it  is  to  be 
shown  that  we  shall  still  have 

ABCD  :  AEFD  :  :  AB  :  AE. 

For  if  not,  the  first  three  terms  continuing 
the  same,  the  fourth  must  be  greater  or  less 
than  AE.     Suppose  it  to  be  greater,  and  that  we  have 

ABCD  :  AEFD  :  :  AB  :  AO. 

Divide  the  line  AB  into  equal  parts,  each  less  than  EO. 
There  will  be  at  least  one  point  I  of  division  between  E  and 
O  :  from  this  point  draw  IK  perpendicular  to  AI :  the  bases 
AB,  AI,  will  be  commensurable,  and  thus,  from  what  is  proved 
above,  we  shall  have 

ABCD  :  AIKD  :  :  AB  :  AI. 

But  by  hypothesis  we  have 

ABCD  :  AEFD  :  :  AB  :  AO. 

In  these  two  proportions  the  antecedents  are  equal ;  hence 
the  consequents  are  proportional  (Book  II.  Prop.  IV.)  ;  and 
we  find 

AIKD  :  AEFD  :  :  AI  :  AO. 

But  AO  is  greater  than  AI ;  hence,  if  this  proportion  is  cor- 
rect, the  rectangle  AEFD  must  be  greater  than  AIKD :  on 
the  contrary,  however,  it  is  less  ;  hence  the  proportion  is  im- 
possible ;  therefore  ABCD  cannot  be  to  AEFD,  as  AB  is  to  a 
line  greater  than  AE. 


EIOB 


n  GEOMETRY. 

Exactly  in  the  same  manner,  it  may  be  shown  that  the  fourth 
term  of  the  proportion  cannot  be  less  than  AE  ;  therefore  it  is 
equal  to  AE. 

Hence,  whatever  be  the  ratio  of  the  bases,  two  rectangles 
ABCD,  AEFD,  of  the  same  altitude,  are  to  each  other  as  their 
bases  AB,  AE. 


PROPOSITION  IV.    THEOREM. 

Any  two  rectangles  are  to  each  other  as  the  products  of  their  bases 
multiplied  by  their  altitudes. 

Let  ABCD,  AEGF,  be  two  rectangles  ;  then  will  the  rect- 
angle, 

ABCD  :  AEGF  :  :  AB.AD  :  AF.AE. 

Having  placed  the  two  rectangles, 
so  that  the  angles  at  A  are  vertical, 
produce  the  sides  GE,  CD,  till  they 
meet  in  H.  The  two  rectangles 
ABCD,  AEHD,  having  the  same  al- 
titude AD,  are  to  each  other  as  their 
bases  AB,  AE  :  in  like  manner  the 
two  rectangles  AEHD,  AEGF,  having  the  same  altitude  AE, 
are  to  each  other  as  their  bases  AD,  AF :  thus  we  have  the 
two  proportions, 

ABCD  :  AEHD  :  :  AB  ;  AE, 
AEHD  :  AEGF  :  :  AD  :  AF. 

Multiplying  the  corresponding  terms  of  these  proportions 
together,  and  observing  that  the  term  AEHD  may  be  omit- 
ted, since  it  is  a  multiplier  of  both  the  antecedent  and  the  con- 
sequent, we  shall  have 

ABCD  :  AEGF  :  :  ABxAD  :  AExAF. 

Scholium,  Hence  the  product  of  the  base  by  the  altitude  may 
be  assumed  as  the  measure  of  a  rectangle,  provided  we  under- 
stand by  this  product,  the  product  of  two  numbers,  one  of 
which  is  the  number  of  linear  units  contained  in  the  base,  the 
other  the  number  of  linear  units  contained  in  the  altitude.  This 
product  will  give  the  number  of  superficial  units  in  the  surface  ; 
because,  for  one  unit  in  height,  there  are  as  many  superficial 
units  as  there  are  linear  units  in  the  base ;  for  two  units  in 
height  twice  as  many ;  for  three  units  in  height,  three  times  as 
many,  &c. 

Still  this  measure  is  not  absolute,  but  relative  :  it  supposes 


IT 

D 

c 

■R 

B 

G 

-A. 

BOOK  IV. 


73 


that  the  area  of  any  other  rectangle  is  computed  m  a  similar 
manner,  by  measuring  its  sides  with  the  same  linear  unit ;  a 
second  product  is  thus  obtained,  and  the  ratio  of  the  two  pro- 
ducts is  the  same  as  that  of  the  rectangles,  agreeably  to  the 
proposition  just  demonstrated. 

For  example,  if  the  base  of  the  rectangle  A  contains  three 
units,  and  its  altitude  ten,  that  rectangle  will  be  represented 
by  the  number  3  x  10,  or  30,  a  number  which  signifies  nothing 
while  thus  isolated  ;  but  if  there  is  a  second  rectangle  B,  the 
base  of  which  contains  twelve  units,  and  the  altitude  seven,  this 
second  rectangle  will  be  represented  by  the  number  12x7  = 
84  ;  and  we  shall  hence  be  entitled  to  conclude  that  the  two 
rectangles  are  to  each  other  as  30  is  to  84  ;  and  therefore,  if 
the  rectangle  A  were  to  be  assumed  as  the  unit  of  measurement 
in  surfaces,  the  rectangle  B  would  then  have  ^^  for  its  absolute 
measure,  in  other  words,  it  would  be  equal  to  ||  of  a  super- 
ficial unit.  .  . 

It  is  more  common  and  more 
simple,  to  assume  the  square  as 
the  unit  of  surface  ;  and  to  se- 
lect that  square,  whose  side  is 
the  unit  of  length.  In  this  case 
the  measurement  which  we  have 

regarded  merely  as  relative,  becomes  absolute  :  the  number  30, 
for  instance,  by  which  the  rectangle  A  was  measured,  now 
represents  30  superficial  units,  or  30  of  those  squares,  which 
have  each  of  their  sides  equal  to  unity,  as  the  diagram  exhibits. 

In  geometry  the  product  of  two  lines  frequently  means  the 
same  thing  as  their  rectangle,  and  this  expression  has  passed 
into  arithmetic,  where  it  serves  to  designate  the  product  of  two 
unequal  numbers,  the  expression  square  being  employed  to 
designate  the  product  of  a  number  multiplied  by  itself. 

The  arithmetical  squares  of  1,  2,  3, 
&c.  are  1,  4,  9,  &c.  So  Hkewise,  the 
geometrical  square  constructed  on  a 
double  line  is  evidently  four  times 
greater  than  the  square  on  a  single  One ; 
on  a  triple  line  it  is  nine  times  great- 
er, &c. 

10 


-A 


74  /  GEOMETRY. 


PROPOSITION  V.    THEOREM. 

The  area  of  any  parallelogram  is  equal  to  the  product  of  its  base 
by  its  altitude. 

For,  the  parallelogram  ABCD  is  equivalent  jF  D         EC 
to  the  rectangle  ABEF,  which  has  the  same 
base  AB,  and  the  same  altitude  BE  (Prop.  I. 
Cor.) :  but  this  rectangle  is  measured  by  AB 


X  BE  (Prop.  IV.  Sch.) ;  therefore,  AB  x  BE  A.  B 

is  equal  to  the  area  of  the  parallelogram  ABCD. 

Cor.  Parallelograms  of  the  same  base  are  to  each  other  as 
their  altitudes ;  and  parallelograms  of  the  same  altitude  are  to 
each  other  as  their  bases :  for,  let  B  be  the  common  base,  and 
C  and  D  the  altitudes  of  two  parallelograms : 

then,        BxC  :  BxD  :  :  C  :  D,  (Book  II.  Prop.  VII.) 

And  if  A  and  B  be  the  bases,  and  C  the  common  altitude, 
we  shall  have 

AxC  :  BxC  :  :  A  :  B. 

And  parallelograms,  generally,  are  to  each  other  as  the  pro- 
ducts of  their  bases  and -altitudes. 


PROPOSITION  VI.    THEOREM 

The  area  of  a  triangle  is  equal  to  the  product  of  its  base  by  half 
its  al^i'^'de. 

For,  the  triangle  ABC  is  half  of  the  par- 
allelogram ABCE,  which  has  the  same  base 
BC,  and  the  same  altitude  AD  (Prop.  II.)  ; 
but  the  area  of  the  paralfelogram  is  equal  to 
BC  X  AD  (Prop.  V.) ;  hence  that  of  the  trian- 
gle must  be  iBC  x  AD,  or  BC  x  \KD. 

Cor.  Two  triangles  of  the  same  altitude  are  to  each  other  as 
their  bases,  and  two  triangles  of  the  same  base  are  to  each 
other  as  their  altitudes.  And  triangles  generally,  are  to  each 
other,  as  the  products  of  their  bases  and  altitudes. 


BOOK  IV.  75 


PROPOSITION  VII.    THEOREM. 

2%e  area  of  a  trapezoid  is  equal  to  its  altitude  multiplied  by  the 
half  sum  of  its  parallel  bases. 

Let  ABCD  be  a  trapezoid,  EF  its  alti- 
tude, AB  and  CD  its  parallel  bases;  then 
will  its  area  be  equal  to  EF  x  i(AB  +  CD). 

Through  I,  the  middle  point  of  the  side 
BC,  draw  KL  parallel  to  the  opposite  side 
AD ;  and  produce  DC  tHl  it  meets  KL. 

In  the  triangles  IBL,  ICK,  we  have  the  side  IB=IC,  by 
construction;  the  angle  LIB  =  CIK;  and  since  CK  and  BL 
are  parallel,  the  angle  IBL=ICK  (BookL  Prop.  XX.  Cor.  2.); 
hence  the  triangles  are  equal  (Book  L  Prop.  VL);  therefore, 
the  trapezoid  ABCD  is  equivalent  to  the  parallelogram  ADKL, 
and  is  measured  by  EF  x  AL. 

But  we  have  AL  =  DK ;  and  since  the  triangles  IBL  and 
KCI  are  equal,  the  side  BL=:CK:  hence,  AB4- CD  =^AL + 
DK=2AL ;  hence  AL  is  the  half  sum  of,  the  bases  AB,  CD ; 
hence  the  area  of  the  trapezoid  ABCD,  is  equal  to  the  altitude 
EF  multiplied  by  the  half  sum  of  the  bases  AB,  CD,  a  result 

which  is  expressed  thus :  ABCD=EF  x  :^5±J^. 

Scholium.  If  through  I,  the  middle  point  of  BC,  the  line  IH 
be  drawn  parallel  to  the  base  AB,  the  point  H  will  also  be  the 
middle  of  AD.  For,  since  the  figure  AHIL  is  a  parallelogram, 
as  also  DHIK,  their  opposite  sides  being  parallel,  we  have 
AH=1L,  and  I)H=IK;  but  since  the  triangles  BIL,  CIK,  are 
equal,  we  already  have  IL=IK;  therefore,  AH=DH. 

It    may  be   observed,  that  the    line    HI=AL   is  equal  to 

i ;  hence  the  area  of  the  trapezoid  may  also  be  ex- 
pressed by  EF  X  HI :  it  is  therefore  equal  to  the  altitude  of  the 
trapezoid  multiplied  by  the  line  which  connects  the  middle 
points  of  its  inclined  sides. 


76  GEOMETRY. 


PROPOSITION  VIII.    THEOREM. 

If  a  line  is  divided  into  two  parts,  the  square  described  on  the 
whole  line  is  equivalent  to  the  sum  of  the  squares  described  on 
the  parts,  together  with  twice  the  rectangle  contained  by  the 
parts. 

Let  AC  be  the  line,  and  B  the  point  of  division ;  then,  is 
AC2  or  (AB  +  BC)2=AB2+BC2+2ABxBC. 

Construct   the   square  ACDE ;   take  AF=    12  H    I) 

AB ;  draw  FG  parallel  to  AC,  and  ^H  par- 
allel to  AE. 

The  square  ACDE  is  made  up  of  four  parts  ; 
the  first  ABIF  is  the  square  described  on  AB, 
since  we  made  AF= AB :  the  second  IDGH  is    K  "Q    C 

the  square  described  on  IG,  or  BC  ;  for  since  we  have  AC  = 
AE  and  AB=AF,  the  difference,  AC — AB  must  be  equal  to 
the  difference  AE — AF,  which  gives  BC=EF  ;  butIG  is  equal 
to  BC,  and  DG  to  EF,  since  the  lines  are  parallel ;  therefore 
IGDH  is  equal  to  a  square  described  on  BC.  And  those  two 
squares  being  taken  away  from  the  whole  square,  there  re- 
mains the  two  rectangles  BCGI,  EFIH,  each  of  which  is  mea- 
sured by  AB  X  BC  :  hence  the  large  square  is  equivalent  to  the 
two  small  squares,  together  with  the  two  rectangles. 

Cor.  If  the  line  AC  were  divided  into  two  equal  parts,  the 
two  rectangles  EI,  IC,  would  become  squares,  and  the  square 
described  on  the  whole  line  would  he  equivalent  to  four  times 
the  square  described  on  half  the  line. 

Scholium,  This  property  is  equivalent  to  the  property  de- 
monstrated in  algebra,  in  obtaining  the  square  of  a  binominal; 
which  is  expressed  thus  : 

(a+6)2=:aH2a6  +  6^ 


PROPOSITION  IX.    THEOREM. 

The  square  described  on  the  difference  of  two  lines,  is  equivalent 
to  the  sum  of  the  squares  described  on  the  lines,  minus  twice 
the  rectangle  contained  by  the  lines. 


BOOK  IV. 


77 


Let  AB  and  BC  be  two  lines,  AC  their  difference ;  then  is 
AC2,  or  (AB— BC)2=:AB2+BC2— 2ABxBC. 

Describe  the  square  ABIF  ;  take  AE    JL      IP  G-      I 

=AC  ;  draw  CG  parallel  to  to  BI,  HK 
parallel  to  AB,  and  complete  the  square 
EFLK. 

The  two  rectangles  GBIG,  GLKD, 
are  each  measured  by  AB  x  BC  ;  take 
them  away  from  the  whole  figure 
ABILKEA,  which  is  equivalent  to 
AB^+BC^  and  there  will  evidently  remain  the  square  ACDE; 
hence  the  theorem  is  true. 

Scholium.    This  proposition  is  equivalent  to  the  algebraical 
formula,  {a-^f=d'-^2ah-^h\ 


K 

E 

D 

A             C 

B 

PROPOSITION  X.    THEOREM. 

The  rectangle  contained  by  the  sum  and  the  difference  of  two 
lines,  is  equivalent  to  the  difference  of  the  squares  of  those 
lines. 


O    I 


Let  AB,  BC,  be  two  lines ;  then,  will 

(AB+BC)  X  (AB— BC)=rAB2— BC2. 

On  AB  and  AC,  describe  the  squares    -p, 
ABIF,  ACDE ;  produce  ABtill  the  pro- 
duced part  BK  is   equal  to  BC  ;  and 
complete  the  rectangle  AKLE. 

The  base  AK  of  the  rectangle  EK, 
is  the  sum  of  the  two  lines  AB,  BC  ;  its 
altitude  AE  is  the  difference  of  the 
same  lines ;  therefore  the  rectangle 
AKLE  is  equal  to  (AB  +  BC)  x  (AB— 
BC).  But  this  rectangle  is  composed  of  the  two  parts  ABHE 
4-BHLK  ;  and  the  part  BHLK  is  equal  to  the  rectangle  EDGF, 
because  BH  is  equal  to  DE,  and  BK  to  EF  ;  hence  AKLE  is 
equal  to  ABHE  +  EDGF.  These  two  parts  make  up  the  square 
ABIF  minus  the  square  DHIG,  which  latter  is  equal  to  a 
square  described  on  BC  :  hence  we  have 

(AB+BC)  X  (AB— BC)-AB2-^BC2 

Scholium.     This  proposition  is  equivalent  to  the  algebraical 
formula,  {a  +  b)x  (a — b)  =a^ — b^. 


E 

h: 

X 

D 

L 

J 

^               ( 

J    I 

1    I 

L 

G* 


78 


OEOMETRY. 


PROPOSITION  XI.     THEOREM. 


The  square  described  on  the  hypothenuse  of  a  right  angled  tri- 
angle is  equivalent  to  the  sum  of  the  squares  described  on  the 

other  two  sides. 


Let  the  triangle  ABC  be  right 
angled  at  A.  Having  described 
squares  on  the  three  sides,  let 
fall  from  A,  on  the  hypothenuse, 
the  perpendicular  AD,  which 
produce  to  E;  and  draw  the 
diagonals  AF,  CH. 

The  angle  ABF  is  made  up 
of  the  angle  ABC,  together  with 
the  right  angle  CBF ;  the  angle 
CBH  is  made  up  of  the  same 
angle  ABC,  together  with  the 
right  angle  ABH  ;   hence  the  -^        e 

angle  ABF  is  equal  to  HBC.  But  we  have  AB=BH,  being 
sides  of  the  same  square ;  and  BF=:BC,  for  the  same  reason  : 
therefore  the  triangles  ABF,  HBC,  have  two  sides  and  the  in- 
cluded angle  in  each  equal ;  therefore  they  are  themselves 
equal  (Book  I.  Prop.  V.). 

The  triangle  ABF  is  half  of  the  rectangle  BE,  because  they 
have  the  same  base  BF,  and  the  same  altitude  BD  (Prop.  II. 
Cor.  1.).  The  triangle  HBC  is  in  like  manner  half  of  the 
square  AH :  for  the  angles  BAC,  BAL,  being  both  right  angles, 
AC  and  AL  form  one  and  the  same  straight  line  parallel  to 
HB  (Book  I.  Prop.  III.) ;  and  consequently  the  triangle  HBC, 
and  the  square  AH,  which  have  the  common  base  BH,  have 
also  the  common  altitude  AB  ;  hence  the  triangle  is  half  of  the 
square. 

The  triangle  ABF  has  already  been  proved  equal  to  the  tri- 
angle HBC  ;  hence  the  rectangle  BDEF,  which  is  double  of 
the  triangle  ABF,  must  be  equivalent  to  the  square  AH,  which 
is  double  of  the  triangle  HBC.  In  the  same  manner  it  may  be 
proved,  that  the  rectangle  CDEG  is  equivalent  to  the  square 
AI.  But  the  two  rectangles  BDEF,  CDEG,  taken  together, 
make  up  the  square  BCQF  :  therefore  the  square  BCGF,  de- 
scribed on  the  hypothenuse,  is  equivalent  to  the  sum  of  the 
squares  ABHL,  ACIK,  described  on  the  two  other  sides  ;  in 
other  words,  BC-r-ABHAC^. 


BOOK  IV.  79 

Cor.  1.  Hence  the  square  of  one  of  the  sides  of  a  right  an- 
gled triangle  is  equivalent  to  the  square  of  the  hypothenuse 
diminished  by  the  square  of  the  other  side  ;  which  is  thus  ex- 
pressed :  AB2=BC2—AC2.  • 

Cor.  2.    It  has  just  been  shown  that  the  square  AH  is  equi- 
valent to  the  rectangle  BDEF ;  but  by  reason  of  the  common 
altitude  BF,  the  square  BCGF  is  to  the  rectangle  BDEF  as  the 
base  BC  is.  to  the  base  BD ;  therefore  we  have 
BC2  :  AB2  :  :  BC  :  BD. 

Hence  the  square  of  the  hypothenuse  is  to  the  square  of  one  of 
the  sides  about  the  right  angle,  as  the  hypothenuse  is  to  the  seg- 
ment adjacent  to  that  side:  The  word  segment  here  denotes 
that  part  of  tlie  hypothenuse,  which  is  cut  off  by  the  perpen- 
dicular let  fall  from  the  right  angle  :  thus  BD  is  the  segment 
adjacent  to  the  side  AB  ;  and  DC  is  the  segment  adjacent  to 
the  side  AG.     We  might  have,  in  like  manner, 

BC2  :  AC2  :  :  BC  :  CD. 

Cor.  3.  The  rectangles  BDEF,  DCGE,  having  Hkewise  the 
same  altitude,  are  to  each  other  as  their  bases  BD,  CD.  But 
these  rectangles  are  equivalent  to  the  squares  AH,  AI ;  there- 
fore we  have  AB^  :  AC^  :  :  BD  :  DO. 

Hence  the  squares  of  the  two  sides  containing  the  right  angle, 
are  to  each  other  as  the  segments  of  the  hypothenuse  which  lie 
^adjacent  to  those  sides. 

Cor.  4.  Let  ABCD  be  a  square,  and  AC  its  ff 
diagonal :  the  triangle  ABC  being  right  an- 
gled and  isosceles,  we  shall  have  AC^— AB^+ 
BC^=2AB^:  hence  the  square  described  on  the 
diagonal  AC,  is  double  of  the  square  described 
on  the  side  AB. 

This  property  may  be  exhibited  more  plainly, 
by  drawing  parallels  to  BD,  through  the  points  A  and  C,  and 
parallels  to  AC,  through  the  points  B  and  D.  A  new  square 
EFGH  will  thus  be  formed,  equal  to  the  square  of  AC.  Now 
EFGH  evidently  contains  eight  triangles  each  equal  to  ABE  ; 
and  ABCD  contains  four  such  triangles :  hence  EFGH  is 
double  of  ABCD. 

Since  we  have  AC"  ;  AB^  :  :  2  :  1  ;  by  extracting  the 
square  roots,  we  shall  have  AC  :  AB  :  :  V2  :  1  ;  hence,  the 
diagonal  of  a  square  is  incommensurable  with  its  side  ;  a  pro- 
perty which  will  be  explained  more  fully  in  another  place. 


80 


GEOMETRY. 


PROPOSITION  XII.    THEOREM. 


In  every  tinangle,  the  square  of  a  side  opposite  an  acute  angle  is 
less  than  the  sum,  of  the  squares  of  the  other  two  sides ^  hy  twice 
the  rectangle  contained  hy  the  base  and  the  distance  from  the 
acute  angle  to  the  foot  of  the  perpendicular  let  fall  from  thb 
opposite  angle  on  the  base,  or  on  the  base  produced. 

Let  ABC  be  a  triangle,  and  AD  perpendicular  to  the  base 
CB ;  then  will  AB2= ACH  BC^— 2BC  x  CD, 

There  are  two  cases. 

First.  When  the  perpendicular  falls  within 
the  triangle  ABC,  we  have  BD==:BC— CD, 
and  consequently  BD^-BCHCD^— 2BC 
xCD  (Prop.  IX.).  Adding  AD^  to  each, 
and  observing  that  the  right  angled  trian- 
gles ABD,  ADC,  give  AD^+BW=AB\  and 
AD2+CD2=AC2,  we  have  AB^^BC^-f  ^ 
AC2— 2BC  X  CD.  ® 

Secondly.  When  the  perpendicular  AD 
falls  without  the  triangle  ABC,  we  have'BD 
=  CD— BC  ;  and  consequently  BD^^CD^^- 
BC^— 2CD  X  BC  (Prop.  IX.).  Adding  AD^ 
to  both,  we  find,  as  before,  AB2=BC2+AC2 
— 2BCxCD. 


PROPOSITION  XIII.    THEOREM. 


In  every  obtuse  angled  triangle,  the  square  of  the  side  opposite  the 
obtuse  angle  is  greater  than  the  sum  of  the  squares  of  the  other 
two  sides  by  twice  the  rectangle  contained  by  the  base  and  the 
distance  from  the  obtuse  angle  to  the  foot  of  the  perpendicular 
let  fall  from  the  opposite  angle  on  the  base  produced. 

Let  ACB  be  a  triangle,  C  the  obtuse  angle,  and  AD  perpen- 
dicular to  BC  produced  ;  then  will  AB2^AC2+BCH2BCx 
CD. 

The  perpendicular  cannot  fall  within  the 
triangle  ;  for,  if  it  fell  at  any  point  such  as 
E,  there  would  be  in  the  triangle  ACE,  the 
right  angle  E,  and  the  obtuse  angle  C,  which 
is  impossible  (Book  L  Prop,  XXV.  Cor.  3.)  ; 


BOOK  IV.  81 

hence  the  perpendicular  falls  without ;  and  we  have  BD=BC 
4-  CD.  From  this  there  results  BD2=BC2  +  CW+ 2BC  x  CD 
(Prop.  VIII.).  Adding  AD^  to  both,  and  reducing  the  sums  as 
in  the  last  theorem,  we  find  AB^^BCH  ACHsBC  x  CD. 

Scholium.  The  right  angled  triangle  is  the  only  one  in  which 
the  squares  described  on  the  two  sides  are  together  equivalent 
to  the  square  described  on  the  tUird ;  for  if  the  angle  contained 
by  the  two  sides  is  acute,  the  sum  of  their  squares  will  be 
greater  than  the  square  of  the  opposite  side ;  if  obtuse,  it  will 
be  less. 

PROPOSITION  XIV.    THEOREM. 

In  any  triangle,  if  a  straight  line  be  dravm  from  the  vertex  to  the 
middle  of  the  base,  twice  the  square  of  this  line,  together  with 
twice  the  square  of  half  the  base,  is  equivalent  to  the  sum  of  the 
squares  of  the  other  two  sides  of  the  triangle. 

Let  ABC  be  any  triangle,  and  AE  a  line  drawn  to  the  mid- 
dle of  the  base  BC  ;  then  will 

2AE2-f2BE2=AB2+AC2. 
On  BC,  let  fail  the  perpendicular  AD. 
Then,  by  Prop.  XII. 

AC2=AEHEC2_2EC  X  ED. 
And  by  Prop.  XIII. 

AB2-AEHEB2+2EB  x  ED.  b"" ED" 

Hence,  by  adding,  and  observing  that  EB  and  EC  are  equal, 
we*^  have. 

AB2 + AC2-2  AE2  +  2EB2. 
Cor.    Hence,  in  every  parallelogram  the  squares  of  the  sides 
are  together  equivalent  to  the  squares  of  the  diagonals. 

For  the  diagonals  AC,  BD,  bisect  each  jg  q 

other  (Book  I.  Prop.  XXXI.) ;  consequently 
the  triangle  ABC  gives 

AB24-BC2=:2AE2+2BE2. 

The  triangle  ADC  gives,  in  like  manner. 

AD^ + DC2r=2  AEH  2DE2. 
Adding  the  corresponding  members  together,  and  observing  that 
BE  and  DE  are  equal,  we  shall  have 

AB2+ADHDCHBC2r=4AE2+4DE2 
But  4AE2  is  the  square  of  2AE,  or  of  AC  ;  4DE^  is  the  squares 
of  BD  (Prop.  VIII.  Cor.) :  hence  the  squares  of  the  sides  are 
together  equivalent  to  the  squares  of  the  diagonals. 


82 


GEOMETRY. 


PROPOSITION  XV.     THEOREM. 


If  a  line  he  drawn  parallel  to  the  base  of  a  triangle^  it  will  divide 
the  other  sides  proportionally. 

Let  ABC  be  a  triangle,  ami  DE  a  straight  line  drawn  par- 
allel to  the  base  BC  ;  then  will 

AD  :  DB  :  :  AE  :  EC. 

Draw  BE  and  DC.  The  two  triangles  BDE, 
DEC  having  the  same  base  DE,  and  the  same 
altitude,  since  both  Iheir  vertices  lie  in  a  line 
parallel  to  the  base,  are  equivalent  (Prop.  II. 
Cor.  2.).  .      ' 

The  triangles  ADE,  BDE,  whose  common 
vertex  is  E,  have  the  same  altitude,  and  are  to 
each  other  as  their  bases  (Prop.  VI.  Cor.)  ; 
hence  we  have 

ADE  :  BDE  :  :  AD  :  DB. 

The  triangles  ADE,  DEC,  whose  common  vertex  is  D,  have 
also  the  same  altitude,  and  are  to  each  other  as  their  bases  ; 
hence 

ADE  :  DEC  :  :  AE  :  EC. 

But  the  triangles  BDE,  DEC,  are  equivalent ;  and  therefore, 
we  have  (Book  II.  Prop.  IV.  Cor.) 

AD  :  DB  :  :  AE  :  EC. 

Cor.  1.  Hence,  by  composition,  we  have  AD  +  DB  :  AD  :  : 
AE  +  EC  :  AE,  or  AB  :  AD  :  :  AC  :  AE  ;  and  also  AB  : 
BD  :  :  AC  :  CE. 

Cor.  2.  If  between  two  straight  lines  AB,  CD,  any  number 
of  parallels  AC,  EF,  GH,  BD,  &c.  be  drawn,  those  straight 
lines  will  be  cut  proportionally,  and  we  shall  have  AE  :  CF :  : 
EG  :  FH  :  GB  :  HD. 

For,  let  O  be  the  point  where  AB  and 
CD  meet.  In  the  triangle  OEF,  the  line 
AC  being  drawn  parallel  to  the  base  EF, 
we  shall  have  OE  :  AE  :  :  OF  :  CF,  or 
OE  ;  OF  :  :  AE  ;  CF.  In  the  triangle 
OGH,  we  shall  likewise  have  OE  :  EG 
:  :  OF  :  FH,brOE  :  OF  :  :  EG  ;  FH. 
And  by  reason  of  the  common  ratio  OE  : 
OF,  those  two  proportions  give  AE  :  CF 
:  :  EG  :  FH.  It  may.be  proved  in  the 
same  manner,  that  EG  :  FH  :  :  GB  :  HD,  and  so  on  ;  hence 
the  lines  AB,  CD,  are  cut  proportionally  by  the  parallels  AC, 
EF,  GH,  &c. 


BOOK  IV.  83 


PROPOSITION  XVI.    THEOREM. 

Conversely,  if  two  sides  of  a  triangle  are  cut  proportionally  by  a 
straight  line,  this  straight  line  will  be  parallel  to  the  third  side. 

In  the  triangle  ABC,  let  the  line  DE  be  drawn,  making 
AD  :  DB  :  t  AE  :  EC  :  then  will  DE  be  parallel  to  BC. 

For,  if  DE  is  not  parallel  to  BC,  draw  DO  paral- 
lel to  it.  Then,  by  the  preceding  theorem,  we  shall 
have  AD  :  PB  :  :  AO  :  OC.  But  by  hypothe- 
sis, we  have  AD  :  DB  :  :  AE  :  EC ;  hence  we 
must  have  AO  :  OC  :  :  AE  ;  EC,or  AO  i  AE 
:  :  OC  :  EC  ;  an  impossible  result,  since  AO,  the 
one  antecedent,  is  less  than  its  consequent  AE, 
and  OC,  the  other  antecedent,  is  greater  than  its 
consequent  EC.  Hence  the  parallel  to  BC,  drawn  from  the 
point  D,  cahnot  differ  from  DE ;  hence  DE  is  that  parallel. 

Scholium,  The  same  conclusion  would  be  true,  if  the  pro- 
portion AB  ;  AD  :  :  AC  :  AE  were  the  proposed  one.  For 
this  proportion  would  give  AB — AD  :  AD  :  :  AC — AE  : 
AE,  or  BD  :  AD  :  :  CE  :  AE. 

PROPOSITION  XVII.     THEOREM. 

The  line  which  bisects  the  vertical  angle  of  a  triangle,  divides  the 
base  into  two  segments,  which  are  proportional  to  the  adjacent 
sides. 

In  the  triangle  ACB,  let  AD  be  drawn,  bisecting  the  angle 
CAB;  then  will 

BD  :  CD  :  :  AB  :  AC. 

Through  the  point  C,  draw  CjE  e 
parallel  to  AD  till  it  meets  BA    K.^. 
produced. 

In  the  triangle  BCE,  the  line  AD 
is  parallel  to  the  base  CE ;  hence 
we  have  the  proportion  (Prop. 
XV.), 

*  BD  :  DC  :  :  AB  :  AE. 

But  the  triangle  ACE  is  isos- 
celes :  for,  since  AD,  CE  are  parallel,  we  have  the  angle  ACE 
=  DAC,  and  the  angle  AEC—BAD  (Book  I.  Prop.  XX.  Cor. 
2  &  3.) ;  but,  by  hypothesis,  DAC=zBAD ;  hence  the  an- 
gle ACErr:AEC,  and  consequently  AEmAC  (Book  I.  Prop. 
XII.).  In  place  of  AE  in  the  above  proportion,  substitute  AC, 
and  we  shall  have  BD  :  DC  :  :  AB  :  AC. 


84     -  GEOMETRY. 


PROPOSITION  XVIII.     THEOREM. 

Tmo.equiangular  triangles  have  their  homologous  sides  propor- 
tional, and  are  similar. 

Let  ABC,  CDE  be  two  triangles  which 
have  their  angles  equal  each  to  each, 
namely,  BAC=r:CDE,  ABC=:DCE  and 
ACB=:  DEC  ;  then  the  homologous  sides, 
or  the  sides  adjacent  to  the  equal  angles, 
will  be  proportional,  so  that  we  shall 
have  BC  :  CE  :  :  AB  :  CD  :  :  AC  : 
DE. 

Place  .the  homologous  sides  BC,  CE  in  the  same  straight 
line  ;  and  produce  the  sides  BA,  ED,  till  they  meet  in  F. 

Since  BCE  is  a  straight  hne,  and  the  angle  BCA  is  equal  to 
CED,  it  follows  that  AC  is  parallel  to  DE  (Book  I.  Prop.  XIX. 
Cor.  2.).  In  like  maimer,  since  the  ai^le  ABC  is  equal  to 
DCE,  the  line  AB  is  parallel  to  DC.  Hence  the  figure  ACDF 
is  a  parallelogram. 

In  the  triangle  BFE,  the  line  AC  is  parallel  to  the  base  FE  ; 
hence  we  have  BC  :  CE  :  :  BA  :  AF  (Prop.  XV.) ;  or  put- 
ting CD  in  the  place  of  its  equal  AF, 

BC  :  CE  :  :  BA  :  CD. 

In  the  same  triangle  BEF,  CD  is  parallel  to  BF  which  may 
be  considered  as  the  base ;  and  we  have  the  proportion 
BC  :  CE  : :  FD  :  DE ;  or  putting  AC  in  the  place  of  its  equal  FD, 

BC  :  CE  :  :  AC  :  DE. 

And  finally,  since  both  these  proportions  contain  the  same 
ratio  BC  :  CE,  we  have 

AC  :  DE  :  :  BA  :  CD. 

Thus  the  equiangular  triangles  BAC,  CED,  have  their  ho- 
mologous sides  proportional.  But  two  figures  are  similar  when 
they  have  their  angles  equal,  each  to  each,  and  their  homolo- 
gous sides  proportional  (Def.  1.)  ;  consequently  the  equiangu- 
lar triangles  BAC,  CED,  are  two  similar  figures. 

Cor.  For  the  similarity  of  two  triangles,  it  is  enough  that 
they  have  two  angles  equal,  each  to  each ;  since  then,  the 
third  will  also  be  equal,  in  both,  and  the  two  triangles  will  be 
equiangular. 


(? 


BOOK  IV.  86 

Scholium,  Observe,  that  in  similar  triangles,  the  homolo- 
gous sides  are  opposite  to  the  equal  angles  ;  thus  the  angle  ACB 
being  equal  to  DEC,  the  side  AB  is  homologous  to  DC  ;  in  like 
manner,  AC  and  DE  are  homologous,  because  opposite  to  the 
equal  angles  ABC,  DCE.  When  the  homologous  sides  are  de- 
termined, it  is  easy  to  form  the  proportions : 

AB  :  DC  :  :  AC  :  DE  :  :  BC  :  CE. 


PROPOSITION  XIX.    THEOREM. 

Two  triangles,  which  have  their  homologous  sides  proportional, 
are  equiangular  and  similar. 

In  the  two  triangles  BAC,  DEF, 
suppose  we  have  BC  :  EF  :  :  AB 
:  DE  :  :  AC  :  DF;  then  will  the 
triangles  ABC,  DEF  have  their  an- 
gles  equal,   namely,  A=D,  B=E, 

At  the  point  E,  make  the  angle 
FEG=B,  and  at  F,  the  angle  EFG=C ;  the  third  G  will  be 
equal  to  the  third  A,  and  the  two  triangles  ABC,  EFG  will  be 
equiangular  (Book  I.  Prop.  XXV.  Cor.  2.).  Therefore,  by  the 
last  theorem,  we  shall  have  BC  :  EF  :  :  AB  :  EG ;  but,  by 
hypothesis,  we  have  BC  :  EF  :  :  AB  :  DE;  hence  EG  =  DE. 
By  the  same  theorem,  we  shall  also  have  BC  :  EF  :  :  AC  ; 
FG;  and  by  hypothesis,  we  have  BC  :  EF  •:  :  AC  :  DF ; 
hence  FG=DF.  Hence  the  triangles  EGF,  DEF,  having  their 
three  sides  equal,  each  to  each,  are  themselves  equal  (Book  I. 
Prop.  X.).  But  by  construction,  the  triangles  EGF "  and 
ABC  are  equiangular ;  hence  DEF  and  ABC  are  also  equian- 
gular and  similar. 

Scholium  1.  By  the  last  two  propositions,  it  appears  that  in 
triangles,  equahty  among  the  angles  is  a  consequence  of  pro- 
portionahty  among  the  sides,  and  conversely  ;  so  that  either  of 
those  conditions  sufficiently  determines  the  similarity  of  two 
triangles.  The  case  is  dilferent  with  regard  to  figures  of 
more  than  three  sides  :  even  in  quadrilaterals,  the  proportion 
between  the  sides  may  be  altered  without  altering  the  angles, 
or  the  angles  may  be  altered  without  altering  the  proportion 
between  the  sides  ;  and  thus  proportionality  among  the  sides 
cannot  be  a  consequence  of  equality  among  the  angles  of  two 
quadrilaterals,  or  vice  versa.     It  is  evident,  for  example,  that 

H 


86 


GEOMETRY. 


by  drawing  EF  parallel  to  BC,  the  angles  of 
the  quadrilateral  AEFD,  are  made  equal  to 
those  of  ABCD,  though  the  proportion  be- 
tween the  sides  is  different ;  and,  in  like  nian- 
ner,  without  changing  the  four  sides  AB,  BC, 
CD,  AD,  we  can  make  the  point  B  approach 
D  or  recede  from  it,  which  will  change  the 
angles. 

Scholium  2.  The  two  preceding  propositions,  which  in  strict- 
ness form  but  one,  together  with  that  relating  to  the  square  of 
the  hypothenuse,  are  the  most  important  aqd  fertile  in  results 
of  any  in  geometry  :  they  are  almost  sufficient  of  themselves 
for  every  application  to  subsequent  reasoning,  and  for  solving 
every  problem.  The  reason  is,  that  all  figures  may  be  divided 
into  triangles,  and  any  triangle  into  two  right  angled  triangles. 
Thus  the  general  properties  of  triangles  include,  by  implica- 
tion, those  of  all  figures. 


PROPOSITION  XX.    THEOREM. 

Two  triangles,  which  have  an  angle  of  the  one  equal  to  an  angle 
of  the  other,  and  the  sides  containing  those  angles  proportional, 
are  similar. 


In  the  two  triangles  ABC,  DEF,  let 
the  angles  A  and  D  be  equal ;  then,  if 
AB  :  DE  :  :  AC  :  DF,  the  two  trian- 
gles will  be  similar. 

Take  AG=DE,  and  draw  GH  paral- 
lel to  BC.  The  angle  AGH  will  be  equal 
to  the  angle  ABC  (Book  I.  Prop.  XX. 
Cor  3.) ;  and  the  triangles  AGH,  ABC,  will  be  equiangular  : 
hence  we  shall  have  AB  :  AG  :  :  AC  :  AH.  But  by  hypo- 
thesis, we  have  AB  :  DE  :  :  AC  :  DF ;  and  by  construction, 
AG=DE:  hence  AH=DF.  The  two  triangles  AGH,  DEF, 
have  an  equal  angle  included  between  equal  sides ;  therefore 
they  are  equal  j  but  the  triangle  AGH  is  similar  to  ABC ;  there- 
fore DEF  is  also  similar  to  ABC. 


BOOK  IV. 


87 


PROPOSITION  XXI.    THEOREM. 


Two  triangles,  which  have  their  homologous  sides  parallel^  or 
perpendicular  to  each  other,  are  similar. 


Let  BAG,  EDF,  be  two  triangles. 

First.  If  the  side  AB  is  parallel  to  DE,  and 
BC  to  EF,  the  angle  ABC  will  be  equal  to 
DEF  (Book  I.  Prop.  XXIV.) ;  if  AC  is  parallel 
to  DF,  the  angle  ACB  will  be  equal  to  DFE, 
and- also  BACtoEDF;  hence  the  triangles 
ABC,  DEF,  are  equiangular;  consequently 
they  are  similar  (Prop.  XVIII.). 


Secondly.  If  the  side  DE  is  perpen- 
dicular to  AB,  and  the  side  DF  to  AC, 
the  two  angles  I  and  H  of  the  quadri- 
lateral AIDH  will  be  right  angles  ;  and 
since  all  the  four  angles  are  together 
equal  to  four  right  angles  (Book  I.  Prop. 
XXVI.  Cor.  l.),the  remaining  two  I  AH, 
IDH,  will  be  together  equal  to  two  right  ^ 
angles.  But  the  two  angles  EDF,  IDH,  are  also  equal  to  two 
right  angles :  hence  the  angle  EDF  is  equal  to  lAH  or  BAC. 
In  like  manner,  if  the  third  side  EF  is  perpendicular  to  the  third 
side  BC,  it  may  be  shown  that  the  angle  DFE  is  equal  to  C,  and 
DEF  to  B  :  hence  the  triangles  ABC,  DEF,  which  have  the 
sides  of  the  one  perpendicular  to  the  corresponding  sides  of  the 
other,  are  equiangular  and  similar. 

Scholium.  In  the  case  of  the  sides  being  parallel,  the  homolo- 
gous sides  are  the  parallel  ones:  in  the  case  of  their  being  per- 
pendicular, the  homologous  sides  are  the  perpendicular  ones. 
Thus  in  the  latter  case  DE  is  homologous  with  AB,  DF  with 
AC,  and  EF  with  BC. 

The  case  of  the  perpendicular  sides  might  present  a  rela- 
tive position  of  the  two  triangles  different  from  that  exhibited 
in  the  diagram.  But  we  might  always  conceive  a  triangle 
DEF  to  be  constructed  within  the  triangle  ABC,  and  such  that 
its  sides  should  be  parallel  to  those  of  the  triangle  compared 
with  ABC  ;  and  then  the  demonstration  given  in  the  text  would 
apply. 


88 


GEOMETRY. 


PROPOSITION  XXII.    THEOREM. 

In  any  triangle,  if  a  line  he  drawn  parallel  to  the  base,  then,  all 
lines  drawn  from  the  vertex  will  divide  the  base  and  the  par- 
allel into  proportional  parts. 

Let  DE  be  parallel  to  the  base  BC,  and 
the  other  lines  drawn  as  in  the  figure ; 
then  will 
DI  :  BF  :  :  IK  :  FG  :  :  KL  :  GH. 

For,  since  DI  is  parallel  to  BF,  the 
triangles  ADI  and  ABF  are  equiangu- 
lar ;  and  we  have  DI  :  BF  :  :  AT  : 
AF ;  and  since  IK  is  parallel  to  FG, 
we  have  in  like  manner  AI  :  AF  :  : 
IK  :  FG;  hence,  the  ratio  AI  i  AF  being  common,  we  shall 
have  DI  :  BF  :  :  IK  ;  FG.  In  the  same  manner  we  shall 
find  IK  :  FG  :  :  KL  :  GH  ;  and  so  with  the  other  segments  : 
hence  the  line  DE  is  divided  at  the  points  I,  K,  L,  in  the  same 
proportion,  as  the  base  BC,  at  the  points  F,  G,  H. 

Cor.  Therefore  if  BC  were  divided  into  equal  parts  at  the 
points  F,  G,  H,  the  parallel  DE  would  also  be  divided  into  equal 
parts  at  the  points  I,  K,  L. 


PROPOSITION  XXIII.    THEOREM. 


If  from  the  right  angle  of  a  right  angled  triangle,  a  perpendicu- 
lar be  let  fall  on  the  hypothenuse ;  then, 

1st.  The  two  partial  triangles  thus  foimed,  will  be  similar  to  each 
other,  and  to  the  whole  triangle. 

2d.  Either  side  including  the  right  angle  will  be  a  mean  propor- 
tional betiveen  the  hypothenuse  and  the  adjacent  segment. 

2d.  The  perpendicular  will  be  a  mean  proportional  between  the 
two  segments  of  the  hypothenuse. 

Let  BAC  be  a  right  angled  triangle,  and  AD  perpendicular 
to  the  hypothenuse  BC. 

First.  The  triangles  BAD  and  BAC 
have  the  common  angle  B,  the  right 
angle  BDA=BAC,  and  therefore  the 
third  angle  BAD  of  the  one,  equal  to 
the  third  angle  C,  of  the  other  (Book 
I.  Prop.  XXV.  Cor  2.) :  hence  those 
two  triangles  are   equiangular    and 


BOOK  IV.  89 

similar.  In  the  same  manner  it  may  be  shown  that  the  trian- 
gles DAC  and  BAG  are  similar ;  hence  all  the  triangles  are 
equiangular  and  similar. 

Secondly,  The  triangles  BAD,  BAG,  being  similar,  their 
homologous  sides  are  proportional.  But  BD  in  the  small  tri- 
angle, and  BA  in  the  large  one,  are  homologous  sides,  because 
they  lie  opposite  the  equal  angles  BAD,  BGA ;  the  hypothe- 
nuse  BA  of  the  small  triangle  is  homologous  with  the  hypo- 
thenuse  BG  of  the  large  triangle  :  hence  the  proportion  BD  : 
BA  :  :  BA  :  BG.  By  the  same  reasoning,  we  should  find 
DG  :  AG  :  ;  AG  :  BG  ;  lience,  each  of  the  sides  AB,  AG,  is 
a  mean  proportional  between  the  hypothenuse  and  the  segment 
adjacent  to  that  side. 

Thirdly.  Since  the  triangles  ABD,  ADG,  are  similar,  by 
comparing  their  homologous  sides,  we  have  BD  :  AD  :  :  AD 
:  DC  ;  hence,  the  perpendicular  AD  is  a  mean  proportional 
between  the  segments  BD,  DG,  of  the  hypothenuse. 

Scholium.  Since  BD  :  AB  :  :  AB  :  BG,  the  product  of  the 
extremes  will  be  equal  to  that  of  the  means,  or  AB^==BD.BG. 
For  the  same  reason  we  have  AG^=DG.BG  ;  therefore  AB^+ 
AG2=BD.BG  +  DG.BG=  (BD  +  DG).BG=BG.BG=BG2;  or 

the  square  described  on  the  hypothenuse  BG  is  equivalent  to 
the  squares  described  on  the  two  sides  AB,  AG.  Thus  we  again 
arrive  at  the  property  of  the  square  of  the  hypothenuse,  by  a 
path  very  different  from  that  which  formerly  conducted  us*  to 
it :  and  thus  it  appears  that,  strictly  speaking,  the  property  of 
the  square  of  the  hypothenuse,  is  a  consequence  of  the  more 
general  property,  that  the  sides  of  equiangular  triangles  are 
proportional.  Thus  the  fundamental  propositions  of  geometry 
are  reduced,  as  it  were,  to  this  single  one,  that  equiangular  tri- 
angles have  their  homologous  sides  proportional. 

It  happens  frequently,  as  in  this  instance,  that  by  deducing 
consequences  from  one  or  more  propositions,  we  are  led  back 
to  some  proposition  already  proved.  In  fact,  the  chief  charac- 
teristic of  geometrical  theorems,  and  one  indubitable  proof  of 
their  certainty  is,  that,  however  we  combine  them  together, 
provided  only  our  reasoning  be  correct,  the  results  we  obtain 
are  always  perfectly  accurate.  The  case  would  be  different, 
if  any  proposition  were  false  or  only  approximately  true  :  it 
would  frequently  happen  that  on  combining  the  propositions 
together,  the  error  would  increase  and  become  perceptible. 
Examples  of  this  are  to  be  seen  in  all  the  demonstrations,  in 
which  the  reductio  ad  absiirdum  is  employed.  In  such  demon- 
strations, where  the  object  is  to  show  that  two  quantities  are 
equal,  we  proceed  by  showing  that  if  there  existed  the  smallest 

HM2 


90 


GEOMETRY. 


inequality  between  the  quantities,  a  train  of  accurate  reason- 
ing would  lead  us  to  a  manifest  and  palpable  absurdity;  from 
which  we  are  forced  to  conclude  that  the  two  quantities  are 
equal. 

Cor.  If  from  a  point  A,  in  the  circumference 
of  a  circle,  two  chords  AB,  AC,  be  drawn  to 
the  extremities  of  a  diameter  BC,  the  triangle 
BAG  will  be  right  angled  at  A  (Book  III.  Prop.  ^  ^  ^ 

XVIII.  Gor.  2.) ;  hence,  first,  the  perpendicular  AD  is  a  mean 
proportional  between  the  two  segments  BD,  DC,  of  the  diameter, 
or  what  is  the  same,  AD^fuBD.DC. 

Hence  also,  in  the  second  place,  the  chord  AB  is  a  mean  pro- 
portional between  the  diameter  BC  and  the  adjacent  segment  BD, 
or,  what  is  the  same,  AB^=BD.BG.  In  like  manner,  we  have 
AC2~CD.BC  ;  hence  AB^  :  AC^  :  :  BD  :  DC  ;  and  com- 
paring AB2  and  AC^,  to  BC^,  we  have  AB^  :  BC^:  :  BD  : 
BC,  and  AC^  :  BC^  :  :  DC  :  BC.  Those  proportions  between 
the  squares  of  the  sides  compared  with  each  other,  or  with  the 
square  of  the  hypothenuse,  have  already  been  given  in  the  third 
and  fourth  corollaries  of  Prop.  XI. 


PROPOSITION  XXIV.    THEOREM. 


Two  triangles  having  an  angle  in  each  equal,  are  to  each  other 
as  the  rectangles  of  the  sides  which  contain  the  equal  angles. 

In  the  two  triangles  ABC,  ADE,  let  the  angle  A  be  equal  to 
the  angle  A ;  then  will  the  triangle 

ABC  :  ADE  :  :  AB.AC  ;  AD.AE. 

Draw  BE.  The  triangles 
ABE,  ADE,  having  the  com- 
mon vertex  E,  have  the  same 
altitude,  and  consequently  are 
to  each  other  as  their  bases 
(Prop.  VI.  Cor.) :  that  is, 

ABE  :  ADE  :  :  AB  •  AD. 

In  like  manner, 

ABC  :  ABE  :  :  AC  :  AE, 

Multiply  together  the  corresponding  terms  of  these  proportions, 
omitting  the  common  term  ABE ;  we  have 

ABC  ;  ADE  :  AB.AC  :  AD.AE. 


BOOK  IV  91 

Cor.  Hence  the  two  triangles  would  be  equivalent,  if  the 
rectangle  AB.AC  were  equal  to  the  rectangle  AD.AE,  or  if 
we  had  AB  :  AD  :  :  AE  :  AC  ;  which  would  happen  if  DC 
were  parallel  to  BE. 


PROPOSITION  XXV.    THEOREM. 

Two  similar  triangles  are  to  each  other  as  the  squares  described 
on  their  homologous  sides. 

Let  ABC,  DEF,  be  two  similar  trian- 
gles, having  the  angle  A  equal  to  D,  and 
the  angle  B=E. 

'  Then,  first,  by  reason  of  the  eqqal  an- 
gles A  and  D,  according  to  the  last  pro- 
position, we  shall  have 

ABC  :  DEF  :  ;  AB.AC  :  DE.DF. 
Also,  because  the  triangles  are  similar, 

AB  :  DE  :  :  AC  :  DF, 
And  multiplying  the  terms  of  this  proportion  by  the  corres- 
ponding terms  of  the  identical  proportion, 

AC  :  DF  :  :  AC  :  DF, 
there  will  result 

AB.AC  :  DE.DF  :  :  AC^  :  DF. 

Consequently, 

ABC  :  DEF  ;  :  AC^  :  DP. 

Therefore,  two  similar  triangles  ABC,  DEF,  are  to  each 
other  as  the  squares  described  on  their  homologous  sides  AC, 
DF,  or  as  the  squiares  of  any  other  two  homologous  sides. 


PROPOSITION  XXVI.    THEOREM. 

Two  similar  polygons  are  composed  of  the  same  number  of  tri^ 
angleSf  similar  each  to  each,  and  similarly  situated. 


92  GEGlVfETRY. 

Let  ABCDE,  FGHIK,  be  two  similar  polygons. 

From  any  angle  A,  in  c 

the  polygon  ABCDE, 
draw  diagonals  AC,  AD 
to  the  other  angles.  From 
the  homologous  angle  F, 
in  the  other  polygon 
-FGHIK,  draw  diagonals 
FH,  FI  to  the  other  an- 
gles. 

These  polygons  being  similar,  the  angles  ABC,  FGH,  which 
are  homologous,  must  be  equal,  and  the  sides  AB,  BC,  must 
also  be  proportional  to  FG,  GH,  that  is,  AB  :  FG  :  :  BC  : 
GH  (Def.  1.).  Wherefore  the  triangles  ABC,  FGH,  have  each 
an  equal  angle,  contained  between  proportional  sides  ;  hence 
they  are  similar  (Prop.  XX.) ;  therefore  the  angle  BC  A  is  equal 
to  GHF.  Take  away  these  equal  angles  from  the  equal  angles 
BCD,  GHI,  and  there  remains  ACD=FHL  But  since  the 
triangles  ABC,  FGH,  are  similar,  we  have  AC  :  FH  :  :  BC  : 
GH ;  and,  since  the  polygons  are  similar,  BC  :  GH  :  :  CD  : 
HI ;  hence  AC  :  FH  :  ;  CD  :  HI.  But  the  angle  ACD,  we 
already  know,  is  equal  to  FHI ;  hence  the  triangles  ACD,  FHI, 
have  an  equal  angle  in  each,  included  between  proportional 
sides,  and  are  consequently  similar  (Prop.  XX.).  In  the  same 
manner  it  might  be  shown  that  all.  the  remaining  triangles  are 
similar,  whatever  be  the  number  of  sides  in  the  polygons  pro- 
posed :  therefore  two  similar  polygons  are  composed  of  the 
same  number  of  triangles,  similar,  and  similarly  situated. 

Scholium.  The  converse  of  the  proposition  is  equally  true  : 
If  two  polygons  are  composed  of  the  same  number  of  triangles 
similar  and  similarly  situated^  those  two  polygons  will  he  similar. 

For,  the  similarity  of  the  respective  triangles  will  give  the 
angles,  ABC  =  FGH,  BCA=GHF,  ACD = FHI :  hence  BCD  = 
GHI,  likewise  CDE=HIK,  &c.  Moreover  we  shall  have 
AB  :  FG  :  :  BC  :  GH  :  :  AC  :  FH  :  :  CD  :  HI, ifec.;  hence 
the  two  polygons  have  their  angles  equal  and  their  sides  pro- 
portional ;  consequently  they  are  similar.  - 


PROPOSITION  XXVII.    THEOREM. 

The  contours  or  perimeters  of  similar  polygons  are  to  each  other 
as  the  homologous  sides  :  and  the  areas  are  to  each  other  as 
the  squares  described  on  those  sides. 


BOOK  IV.  98 

First.  Since,  by  the 
nature  of  similar  figures, 
we  have  AB  :  FG  :  : 
BC  :  GH  ::CD  :  HI, 
&c.  we  conclude  from 
this  series  of  equal  ratios 
that  the  sum  of  the  ante- 
cedents AB  +  BC  +  CD, 
&c.,  which  makes  up  the  perimeter  of  the  first  polygon,  is  to 
the  sum  of  the  consequents  FG+GH  +  HI,  &c.,  which  makes 
up  the  perimeter  of  the  second  polygon,  as  any  one  antecedent 
is  to  its  consequent ;  and  therefore,  as  the  side  AB  is  to  its  cor- 
responding side  FG  (Book  II.  Prop.  X.). 

Secondly.  Since  the  triangles  ABC,  FGH  are  similar,  we 
shall  have  the  triangle  ABC  :  FGH  :  :  AC^  :  FH^  (Prop. 
XXV.) ;  and  in  like  manner,  from  the  similar  triangles  ACD, 
FHI,  we  shall  have  ACD  :  FHI  :  :  AC^  :  FH^;  therefore,  by 
reason  of  the  common  ratio,  AC^  :  FH^  we  have 

ABC  :  FGH  :  :  ACD  :  FHI. 
By  the  same  mode  of  reasoning,  we  should  find 

ACD  :  FHI  :  :  ADE  :  FIK; 
and  so  on,  if  there  were  more  triangles.  And  from  this  series 
of  equal  ratios,  we  conclude  that  the  sum  of  the  antecedents 
ABC  + ACD  + ADE,  or  the  polygon  ABCDE,  is  to  the  sum  of 
the  consequents  FGH  +  FHI  +  FIK,  or  to  the  polygon  FGHIK, 
as  one  antecedent  ABC,  is  to  its  consequent  FGH,  or  as  AB^ 
is  to  FG^  (Prop.  XXV;)  ;  hence  the  areas  of  similar  poly- 
gons are  to  each  other  as  the  squares  described  on  the  homolo- 
gous sides. 

Cor^  If  three  similar  figures  were  constructed,  on  the 
three  sides  of  a  right  angled  triangle,  the  figure  on  the  hypo- 
thenuse  would  be  equivalent  to  the  sum  of  the  other  two  :  for 
the  three  figures  are  proportional  to  the  squares  of  their 
homologous  sides ;  but  the  square-pf  the  hypothenuse  is 
equivalent  to  the  sum  of  the  squares  of  the  two  other  sides ; 
hence,  &c. 


PROPOSITION  XXVIII.    THEOREM. 

The  segments  of  two  chords,  which  intersect  each  other  in  a  circle, 
are  reciprocally  proportional. 


94 


GEOMETRY. 


Let  the  chords  AB  and  CD  intersect  at  O :  then  will 
AO  :  DO  :  :  OC  :  OB. 

Draw  AC  and  BD.  In  the  triangles  ACO, 
BOD,  the  angles  at  O  are  equal,  being  verti- 
cal ;  the  angle  A  is  equal  to  the  angle  D,  be- 
cause both  are  inscribed  in  the  same  segment 
(Book  III.  Prop.  XVIII.  Cor.  1.) ;  for  the  same 
reason  the  angle  CrrB ;  the  triangles  are  there- 
fore similar,  and  the  homologous  sides  give  the  proportion 
AO  :  DO  :  :  CO  :  OB. 

Cor.  Therefore  AO.OB=DO.CO:  hence  th^  rectangle 
under  the  two  segments  of  the  one  chord  is  equal  to  the  rect- 
angle under  the  two  segments  of  the  other. 


PROPOSITION  XXIX.    THEOREM. 


If  from  the  same  point  without  a  circle,  two  secants  he  drawn 
terminating  in  the  concave  arc,  the  whole  secants  will  he  recip- 
rocally proportional  to  their  external  segments. 

Let  the  secants  OB,  OC,  be  drawn  from  the  point  O  : 
then  will 

OB  :  OC  :  :  OD  :  OA. 
For,  drawing  AC,  BD,  the  triangles  OAC, 
OBD  have  the  angle  O  common  ;  likewise  the 
angle  B=C  (Book  III.  Prop.  XVIII.  Cor.  1.); 
these  triangles  are  therefore  similar;  and  their 
homologous  sides  give  thje  proportion, 
OB  :  OC  :  :  OD  :  OA. 

Cor.     Hence  the  rectangle  OA.OB  is  equal 
to  the  rectangle  OC.OD.  .  ^ 

Scholium.  This  proposition,  it  may  be  observed,  bears  a 
great  analogy  to  the  preceding,  and  differs  from  it  only  as  the 
two  chords  AB,  CD,  instead  of  intersecting  each  other  within, 
cut  each  other  without  the  circle.  The  following  proposition 
may  also  be  regarded  as  a  particular  case  of  the  proposition 
just  demonstrated. 


BOOK  IV. 


95 


PROPOSITION  XXX.    THEOREM. 


If  from  the  same  point  without  a  circle,  a  tangent  and  a  secant 
he  drawn,  the  tangent  will  he  a  mean  proportional  hetween  the 
secant  and  its  external  segment. 

From  the  point  O,  let  the  tangent  OA,  and  the  secant  OC  be 
be  drawn  ;  then  will 

OC  :  OA  :  :  OA  :  OD,  or  0A2==0G.0D. 
For,  drawing  AD  and  AC,  the  triangles  O 
OAD,  OAC,  have  the  angle  O  common ;  also 
the  angle  OAD,  formed  by  a  tangent  and  a 
chord,  has  for  its  measure  half  of  the  arc  AD 
(Book  III.  Prop.  XXI.)  ;  and  the  angle  C  has 
the  same  measure :  hence  the  angle  OAD= 
C ;  therefore  the  two  triangles  are  similar, 
and  we  have  the  proportion  OC  :  OA  ;  : 
AO  :  OD,  which  gives  OA^=OC.OD. 


PROPOSITION  XXXI.  •  THEOREM. 

If  either  angle  of  a  triangle  he  bisected  hy  a  line  terminating  in 
the  opposite  side,  the  rectangle  of  the  sides  including  the  bi- 
sected angle,  is  equivalent  to  the  square  of  the  bisecting  line 
together  with  the  rectangle  contained  by  the  segments  of  the 
third  side. 


In  the  triangle  BAC,  let  AD  bisect  the  angle  A ;  then  will 
AB.AC=AD2-fBD.DC. 

Describe  a  circle  through  the  three  points 
A,  B,  C  ;  produce  AD  till  it  meets  the  cir- 
cumference, and  draw  CE. 

The  triangle  BAD  is  similar  iq  the  trian- 
gle EAC  ;  for,  by  hypothesis,  tlie  angle 
BAD=EAC;  also  the  angle  B=E,  since 
they  are  both  measured  by  half  of  the  arc 
AC  ;  hence  these  triangles  are  similar,  and 
the  homologous  sides  give  the  proportion  BA  :  AE  :  :  AD  : 
AC  ;  hence  BA.AC=AE.AD  ;  but  AE=AD  +  DE,  and  multi- 
plying each  of  these  equals  by  AD,  we  have  AE.AD=AD^+ 
AD.DE;  now  AD.DE=BD.DC  (Prop.  XXVIII.);  hence, 
finally, 

BA.AC=AD2+BD.DC. 


GEOMETRY. 


PROPOSITION  XXXIL    THEOREM. 

In  every  triangle,  the  rectangle  contained  by  two  sides  is  equiva- 
lent to  the  rectangle  contained  hy  the  diameter  of  the  circum- 
scribed circle f  and  the  perpendicular  let  fall  upon  the  third- 
side. 


In  the  triangle  ABC,  let  AD  be  drawn  perpendicular  to  BC  ; 
and  let  EC  be-  the  diameter  of  the  circumscribed  circle  ;  then 
will 

AB.AC=:AD.CE. 

For,  drawing  AE,  the  triangles  ABD, 
AEC,  are  right  angled,  the  one  at  D,  the 
other  at  A:  also  the  angle  B=E ;  these  tri- 
angles are  therefore  similar,  and  they  give 
the  proportion  AB  :  CE  :  :  AD  :  AC ;  and 
hence  AB.AC=CE.AD. 

Cor,  If  these  equal  quantities  be  multiplied  by  the  same 
quantity  BC,  there  will  result  AB.AC.BC=CE.AD.BC  ;  now 
AD.BC  is  double  of  the  area  of  the  triangle  (Prop.  VL) ;  there- 
fore the  product  of  three  sid^s  of  a  triangle  is  equal  to  its  area 
multiplied  by  twice  the  diameter  of  the  circumscribed- circle.  " 

The  product  of  three  lines  is  sometimes  called  a  solid,  for  a 
reason  that  shall  be  seen  afterwards.  Its  value  is  easily  con- 
ceived, by  imagining  that  the  lines  are  reduced  into  numbers, 
and  multiplying  these  numbers  together. 

Scholium.  It  may  also  be  demonstrated,  that  the  area  of 
a  triangle  is  equal  to  its  perimeter  multiplied  by  half  the  radius 
of  the  inscribed  circle.    . 

For,  the  triangles  AOB,  BOC, 
AOC,  which  have  a  common 
vertex  at  O,  have  for  their  com- 
mon altitude  the  radius  of  the 
inscribed  circle  ;  hence  the  sum 
of  these  triangles  will  be  equal 
to  the  sum  of  the  bases  AB,  BC,  . 
AC,  multiplied  by  half  the  radius 
OD ;  hence  the  area  of  the  triangle  ABC  is  equal  to  the 
perimeter  multiplied  by  half  the  radius  of  the  inscribed  circle. 


BOOK  IV. 


97 


PROPOSITION  XXXIII.    THEOREM. 

In  every  quadrilateral  inscribed  in  a  circle^  the  rectangle  of  the 
two  diagonals  is  equivalent  to  the  sum  of  the  rectangles  of  the 
opposite  sides. 

In  the  quadrilateral  ABCD,  we  shall  have 
AC.BD = AB.CD + AD.BC. 

Take  the  arc  CO— AD,  and  draw  BO 
meeting  the  diagonal  AC  in  I. 

The  angle  ABD=CBI,  since  the  one 
has  for  its  measure  half  of  the  arc  AD, 
and  the  other,  half  of  CO,  equal  to  AD  ; 
the  angle  ADB=:BCI,  because  they  are 
both  inscribed  in  the  same  segment 
AOB  ;  hence  the  triangle  ABD  is  similar 
to  the  triangle  IBC,  and  we  have  the 
proportion  AD  :  CI  :  :  BD  :  BC ;  hence  AD.BC =CI.BD. 
Again,  the  triangle  ABl  is  similar  to  the  triangle  BDC  ;  for  the 
arc  AD  being  equal  to  CO,  if  OD  be  added  to  each  of  them, 
we  shall  have  the  arc  AO==DC  ;  hence  the  angle  ABI  is  equal 
to  DBC  ;  also  the  angle  BAI  to  BDC,  because  they  are  in- 
scribed in  the  same  segment ;  hence  the  triangles  ABI,  DBC, 
are  similar,  and  the  homologous  sides  give  the  proportion  AB  : 
BD  :  :  AI  :  CD ;  hence  AB.CD=AI.BD. 

Adding  the  two  results  obtained,  and  observing  that 

AI.BD  +  CI.BD  =  (AI  +  CI).BDr=AC.BD, 
we  shall  have 


AD.BC +AB.CD=AC.BD. 


113 


GEOMETRY. 


PROBLEMS  RELATING  TO  THE  FOURTH  BOOK. 


PROBLEM  I. 

To  divide  a  given  straight  line  into  any  number  of  equal  parts, 
or  into  parts  proportional  to  given  lines. 

First,  Let  it  be  proposed  to  divide  the  line 
AB  into  five  equal  parts.  Through  the  ex- 
tremity A,  draw  the  indefinite  straight  line 
AG ;  and  taking  AC  of  any  magnitude,  apply 
it  five  times  upon  AG  ;  join  the  last  point 
of  division  G,  and  the  extremity  B,  by  the 
straight  line  GB  ;  then  draw  CI  parallel  to 
GB :  AI  will  be  the  fifth  part  of  the  line 
AB ;  and  thus,  by  applying  AI  five  times 
upon  AB,  the  line  AB  will  be  divided  into 
five  equal  parts. 

For,  since  CI  is  parallel  to  GB,  the  sides  AG,  AB,  are  cut 
proportionally  in  C  and  I  (Prop.  XV.).  But  AC  is  the  fifth 
part  of  AG,  hence  AI  is  the  fifth  part  of  AB, 

Secondly.  Let  it  be  pro- 
posed to  divide  the  line  AB 
into  parts  proportional  to 
the  given  lines  P,  Q,  R. 
Through  A,  draw  the  indefi- 
nite line  AG  ;  make  AC  = 
P,  CD=Q,  DE=R;  join 
the  extremities  E  and  B  ; 
and    through   the  points  C, 

D,  draw  CI,  DF,  parallel  to  EB ;  the  line  AB  will  be  divided 
into    parts  AI,  IF,  FB,  proportional   to    the  given  lines  P, 

Q,R. 

For,  by  reason  of  the  para..cis  CI,  DF,  EB,  the  parts  AI, 
IF,  FB,  are  proportional  to  the  parts  AC,  CD,  DE  ;  and  by 
construction,  these  are  equal  to  the  given  lines  P,  Q,  R. 


BOOK  IV.  99 

PROBLEM  II. 

To  find  a  fourth  proportional  to  three  given  lines.  A,  B,  C. 

Draw  the  two  indefi- 
nite lines  DE,  DF,  form- 
ing any  angle  with  each 
other.  Upon  DE  take 
DA=A,  and  DB=:B; 
upon  DF  take  DC=C  ; 
draw  AC  ;  and  through 
the  point  B,  draw  BX 
parallel  to  AC  ;  DX  will  be  the  fourth  proportional  required  ; 
for,  since  BX  is  parallel  to  AC,  we  have  the  proportion 
DA  :  DB  :  :  DC  :  DX ;  now  the  first  three  terms  of  this  pro- 
portion are  equal  to  the  three  given  lines  :  consequently  DX  is 
the  fourth  proportional  required. 

Cor.  A  third  proportional  to  two  given  lines  A,  B,  may  be 
found  in  the  same  manner,  for  it  will  be  the  same  as  a  fourth 
proportional  to  the  three  lines  A,  B,  B. 

PROBLEM  III. 
To  find  a  mean  proportional  between  two  given  lines  A  and  B. 

Upon  the  indefinite  line  DF,  take 
DE=A,  and  EF=:B  ;  upon  the  whole 
line  DF,  as  a  diameter,  describe  the 
semicircle  DGF ;  at  the  point  E, 
erect  upon  the  diameter  the  perpen- 
dicular EG  meeting  the  circumfe- 
rence in  G ;  EG  will  be  the  mean 
proportional  required. 

For,  the  perpendicular  EG,  let  fall  from  a  point  in  the  cir- 
cumference upon  the  diameter,  is  a  mean  proportional  between 
DE,  EF,  the  two  segments  of  the  diameter  (Prop.  XXIII. 
Cor.) ;  and  these  segments  are  equal  to  the  given  lines  A 
andB. 

PROBLEM  IV. 

To  divide  a  given  line  into  two  partSy  such  that  the  greater  part 
shall  be  a  mean  proportional  between  the  whole  line  and  the 
other  part. 


lt)0  GEOMETRY. 

Let  AB  be  the  given  line. 

At  the  extremity  B  of  the  line 
AB,  erect  the  perpendicular  BC 
equal  to  the  half  of  AB  ;  from  the 
point  C,  as  a  centre,  with  the  ra- 
dius CB,  describe  a  semicircle  ; 

draw  AC  cutting  the  circumfe-     

rence  in  D  ;  and  take  AF=AD  :     ^  E         B 

the  line  AB  will  be  divided  at  the  point  F  in  the  manner  re- 
quired ;  that  is,  we  shall  have  AB  :  AF  :  :  AF  :  FB. 

For,  AB  being  perpendicular  to  the  radius  at  its  extremity, 
is  a  tangent ;  and  if  AC  be  produced  till  it  again  meets  the 
circumference  in  E,  we  shall  have  AE  ;  AB  :  :  AB  :  AD 
(Prop.  XXX.) ;  hence,  by  division,  AE— AB  :  AB  :  :  AB— 
AD  :  AD.  But  since  the  radius  is  the  half  of  AB,  the  diame- 
ter DE  is  equal  to  AB,  and  consequently  AE — AB=AD=AF  ; 
also,  because  AF=AD,  we  have  AB — AD=FB ;  hence 
AF  :  AB  :  :  FB  :  AD  or  AF ;  whence,  by  exchanging  the 
extremes  for  the  means,  AB  :  AF  :  :  AF  :  FB. 

SchAium.  This  sort  of  division  of  the  line  AB  is  called  di- 
vision in  extreme  and  mean  ratio  :  the  use  of  it  will  be  per- 
ceived in  a  future  part  of  the  work.  It  may  further  be 
observed,  that  the  secant  AE  is  divided  in  extreme  and  mean 
ratio  at  t(he  point  D  ;  for,  since  AB=DE,  we  have  AE  :  DE 
:  :  DE  :  AD. 


PROBLEM  V. 

Through  a  given  pointy  in  a  given  angle,  to  draw  a  line  so  that 
the  segments  comprehended  between  the  point  and  the  two  sides 
of  the  angle,  shall  be  equal. 

Let  BCD  be  the  given  angle,  and  A  the  given  point. 

Through  the  point  A,  draw  AE  paral- 
lel to  CD,  make  BE=CE,  and  through 
the  points  B  and  A  draw  BAD  ;  this  will 
be  the  line  required. 

For,  AE  being  parallel  to  CD,  we  have 
BE  :  EC  :  :  BA  :  AD ;  but  BE=EC  ; 
therefore  BA=AD. 


BOOK  IV. 

PROBLEM  VI. 


101 


To  describe  a  square  that  shall  he  equivalent  to  a  given  parallelo- 
gram, or  to  a  given  triangle. 

First.     Let  ABCD  be  X  Y 

the  given  parallelogram, 
AB  its  base,  DE  its  alti- 
tude :  between  AB  and 
DE  find  a  mean  propor- 
tional XY ;  then  will  the 
square  described  upon 
XY  be  equivalent  to  the  parallelogram  ABCD. 

For,  by  construction,  AB  :  XY  :  :  XY  :  DE ;  therefore, 
XY^= AB.DE ;  but  AB.DE  is  the  measure  of  the  parallelogram, 
and  XY^  that  of  the  square  ;  consequently,  they  are  equiva- 
lent 

Secondly.  Let  ABC  be  the 
given  triangle,  BC  its  base, 
AD  its  altitude  :  find  a  mean 
proportional  between  BC  and 
the  half  of  AD,  and  let  XY  be 
that  mean ;  the  square  de- 
scribed upon  XY  will  be  equi- 
valent to  the  triangle  ABC. 

For,  since  BC  :  XY  :  :  XY  :  ^AD,  it  follows  that  XY2= 
BC.^AD  ;  hence  the  square  described  upon  XY  is  equivalent 
to  the  triangle  ABC. 


PROBLEM  VII. 


Upon  a  given  line,  to  describe  a  rectangle  that  shall  be  equiva- 
lent to  a  given  rectangle. 

Let  AD  be  the  line,  and  ABFC  the  given  rectangle. 

Find  a  fourth  propor- 
tional  to  the  three  lines 
AD,  AB,  AC,  and  let  AX 
be  that  fourth  propor- 
tional ;  a  rectangle  con- 
structed with  the  lines 
AD  and  AX  will  be  equi- 
valent to  the  rectangle  ABFC. 

For,  since  AD  :  AB  :  :  AC  :  AX,  it  follows  that  AD. AX = 
AB.AC  ;  hence  the  rectangle  ADEX  is  equivalent  to  the  rect- 
angle ABFC. 


im^      /  "    *  *  '  '     GEOMETRY. 

PROBLEM  VIII. 

To  find  two  lines  whose  ratio  shall  be  the  same  as  the  ratio  of 
two  rectangles  contained  by  given  lines. 

Let  A.B,  CD,  be  the  rectangles  contained  by  the  given  lines 
A,  B,  C,  andD. 

Find  X,  a  fourth  proportional  to  the  three      j^    

lines  B,  C,  D  ;  then  will  the  two  lines  A  and 

X  have  the  same  ratio  to  each  other  as  the      _  ^ 


rectangles  A.B  and  CD.  ^  • ' 

For,  since  B  :  C  :  :  D  :  X,  it  follows  that      ^  i -* 

CD=B.X ;  hence  A.B  :  CD  :  :  A.B  :  B.X     Xi » 

:  :  A  :  X. 

Cor,     Hence  to  obtain  the  ratio  of  the  squares  described 

upon  the  given  lines  A  and  C,  find  a  third  proportional  X  to 

the  lines  A  and  C,  so  that  A  :  C  :  :  C  :  X ;  you  will  then 

have 

A.X=:C2,  or  A2.X=zA.C2 ;  hen6e 
A2  :  C2  :  :  A  :  X. 

PROBLEM  IX. 

To  find  a  triangle  that  shall  be  equivalent  to  a  given  polygon. 

Let  ABCDE  be  the  given  polygon. 
Draw  first  the  diagonal  CE  cutting  off 
the  triangle  CDE ;  through  the  point 
D,  draw  DF  parallel  to  CE,  and  meet- 
ing AE  produced  ;  draw  CF :  the  poly- 
gon ABCDE  will  be  equivalent  to  the 
polygon  ABCF,  which  has  one  side 
less  than  the  original  polygon. 

For,  the  triangles  CDE,  CFE,  have  the  base  CE  common^ 
they  have  also  the  same  altitude,  since  their  vertices  D  and  F,. 
are  situated  in  a  line  DF  parallel  to  the  base  :  these  triangles  are 
therefore  equivalent  (Prop.  II.  Cor.  2.).  Add  to  each  of  them 
the  figure  ABCE,  and  there  will  result  the  polygon  ABCDE, 
equivalent  to  the  polygon  ABCF. 

The  angle  B  may  in  like  manner  be  cut  off,  by  substituting 
for  the  triangle  ABC  the  equivalent  triangle  AGC,  and  thus 
the  pentagon  ABCDE  will  be  changed  into  an  equivalent  trir 
angle  GCF. 

The  same  process  may  be  applied  to  every  other  figure  ; 
for,  by  successively  diminishing  the  number  of  its  sides,  one 
being  retrenched  at  each  step  of  the  process,  the  equivalent 
triangle  will  at  last  be  found. 


BOOK  IV.  103 

Scholium.  We  have  already  seen  that  every  triangle  may 
be  changed  into  an  equivalent  square  (Prob.  VI.) ;  and  thus  a 
square  may  always  be  found  equivalent  to  a  given  rectilineal 
figure,  which  operation  is  called  squaring  the  rectilineal  figure, 
or  finding  the  quadrature  of  it. 

The  problem  of  the  quadrature  of  the  circle,  consists  in  find- 
ing a  square  equivalent  to  a  circle  whose  diameter  is  given. 


PROBLEM  X. 

To  find  the  side  of  a  square  which  shall  be  equivalent  to  the  sum 
or  the  difference  of  two  given  squares. 

Let  A  and  B  be  the  sides  of  the 
given  squares. 

First.  If  it  is  required  to  find  a 
square  equivalent  to  the  sum  of 
these  squares,  draw  the  two  indefi- 
nite lines  ED,  EF,  at  right  angles 
to  each  other;  take  ED = A,  and 
EG=B ;  draw  DG :  this  will  be  the  side  of  the  square  re- 
quired. 

For  the  triangle  DEG  being  right  angled,  the  square  de- 
scribed upon  DG  is  equivalent  to  the  sum  of  the  squares  upon 
ED  and  EG. 

Secondly.  If  it  is  required  to  find  a  square  equivalent  to  the 
difference  of  the  given  squares,  form  in  the  same  manner  the  right 
angle  FEH ;  take  GE  equal  to  the  shorter  of  the  sides  A  and 
B ;  from  the  point  G  as  a  centre,  with  a  radius  GH,  equal  to 
the  other  side,  describe  an  arc  cutting  EH  in  H  :  the  square 
described  upon  EH  will  be  equivalent  to  the  difference  of  the 
squares  described  upon  the  lines  A  and  B. 

For  the  triangle  GEH  is  right  angled,  the  hypothenuse 
GH=A,  and  the  side  GE=B;  hence  the  square  described 
upon  EH,  is  equivalent  to  the  difference  of  the  squares  A 
andB. 

Scholium.  A  square  may  thus  be  found,  equivalent  to  the 
sum  of  any  number  of  squares  ;  for  a  similar  construction  which 
reduces  two  of  them  to  one,  will  reduce  three  of  them  to  two, 
and  these  two  to  one,  and  so  of  others.  It  would  be  the  same, 
if  any  of  the  squares  were  to  be  subtracted  from  the  sura  oi 
the  others. 


lQ4t 


GEOMETRY. 

PROBLEM  XI, 


To  find  a  square  which  shall  he  to  a  given  square  as  a  given 
line  to  a  given  line. 


Let  AC  be  the  given  D 
square,  and  M  and  N  the 
given  hnes. 

Upon  the  indefinite 
line  EG,  take  EF=M, 
and  FG=N  ;  upon  EG 
as  a  diameter  describe  A. 


a  semicircle,  and  at  the  point  F  erect  the  perpendicular  FH. 
From  the  point  H,  draw  the  chords  HG,  HE,  which  produce 
indefinitely :  upon  the  first,  take  HK  equal  to  the  side  AB  of 
the  given  square,  and  through  the  point  K  draw  KI  parallel  to 
EG ;  HI  will  be  the  side  of  the  square  required. 

For,  by  reason  of  the  parallels  KI,  GE,  we  have  HI  :  HK 
:  :  HE  :  HG;  hence,  HP  :  HK^  ;  :  HE^  :  HG^:  but  in  the 
right  angled  triangle  EHG,  the  square  of  HE  is  to  the  square 
of  HG  as  the  segment  EF  is  to  the  segment  FG  (Prop.  XI. 
Cor.  3.),  or  as  M  is  to  N ;  hence  HP  :  HK^  :  ;  M  :  N.  But 
HK=AB  ;  therefore  the  square  described  upon  HI  is  to  the 
square  described  upon  AB  as  M  is  to  N. 

PROBLEM  XIL 

Upon  a  given  line,  to  describe  a  polygon  similar  to  a  given 

polygon. 

Let  FG  be  the  given 
line,  and  AEDCB  the 
given  polygon. 

In  the  given  polygon, 
draw  the  diagonals  AC, 
AD ;  at  the  point  F 
make  the  angle  GFH=: 
BAC,  and  at  the  point 
G  the  angle  FGH=ABC  ;  the  lines  FH,  GH  will  cut  each 
other  in  H,  and  FGH  will  be  a  triangle  similar  to  ABC.  In 
the  same  manner  upon  FH,  homologous  to  AC,  describe  the 
triangle  FIH  similar  to  ADC  ;  and  upon  FI,  homologous  to  AD, 
describe  the  triangle  FIK  similar  to  ADE.  The  polygon 
FGHIK  will  be  similar  to  ABCDE,  as  required. 

For,  these  two  polygons  are  composed  of  the  same  number 
of  triangles,  which  are  similai'  and  similarly  situated  (Prop. 
XXVL  Sch.). 


BOOK  IV.  105 


PROBLEM  XIII. 

T\uo  similar  figures  being  given,  to  describe  a  similar  figure 
which  shall  be  equivalent  to  their  sum  or  their  difference. 

Let  A  and  B  be  two  homologous  sides  of  the  given  figures. 

Find  a  square  equivalent  to  the 
sum  or  to  the  difference  of  the 
squares  described  upon  A  and  B  ; 
let  X  be  the  side  of  that  square  ; 
then  will  X  in  the  figure  required, 
be  the  side  which  is  homologous 
to  the  sides  A  and  B  in  the  given 
figures.  The  figure  itself  may  then 
be  constructed  on  X,  by  the  last  problem. 

For,  the  similar  figures  are  as  the  squares  of  their  homolo- 
gous sides ;  now  the  square  of  the  side  X  is  equivalent  to  the 
sum,  or  to  the  d.ierence  of  the  squares  described  upon  the 
homologous  sides  A  and  B  ;  therefore  the  figure  described  upon 
the  side  X  is  equivalent  to  the  sum,  or  to  the  difference  of  the 
similar  figures  described  upon  the  sides  A  and  B. 


PROBLEM  XIV. 

To  describe  a  figure  similar  to  a  given  figure,  and  bearing  to  it 
the  given  ratio  of  M  to  N. 


Let  A  be  a  side  of  the  given  figure,  X 
the  homologous  side  of  the  figure  required. 
The  square  of  X  must  be  to  the  square  of 
A,  as  M  is  to  N  :  hence  X  will  be  found  by 
(Prob.  XL),  and  knowing  X,  the  rest  will  be 
accompKshed  by  (Prob.  XIL). 


14 


106 


GEOMETRY. 


PROBLEM  XV. 


To  construct  a  figure  similar  to  the  figure  P,  and  equivalent  to 
the  figure  Q. 

Find  M,  the  side  of  a  square 
equivalent  to  the  figure  P,  and 
N,  the  side  of  a  square  equiva- 
lent to  the  figure  Q.  Let  X  be 
a  fourth  proportional  to  the  three 
given  lines,  M,  N,  AB  ;  upon 
the  side  X,  homologous  to  AB, 
describe  a  figure  similar  to  the  figure  P  ;  it  will  also  be  equiva- 
lent to  the  figure  Q. 

For,  calling  Y  the  figure  described  upon  the  side  X,  we  have 
P  :  Y  :  :  AB2  :  X^ ;  but  by  construction,  AB  :  X  :  :  M  :  N, 
or  AW  :  X^  :  :  M^  :  N2 ;  hence  P  :  Y  :  :  M^  :  N'^.  But  by 
construction  also,  M2=P  and  N2=Q;  therefore  P  :  Y  :  :  P  : 
Q;  consequently  Y=Q;  hence  the  figure  Y  is  similar  to  the 
figure  P,  and  equivalent  to  the  figure  Q. 


PROBLEM  XVI. 


To  construct  a  rectangle  equivalent  to  a  given  square^  and  having 
the  sum  of  its  adjacent  sides  equal  to  a  given  line. 

Let  C  be  the  square,  and  AB  equal  to  the  sum  of  the  sides 
of  the  required  rectangle. 

Upon  AB  as  a  diame- 
ter, describe  a  semicir- 
cle ;  draw  the  line  DE 
parallel  to  the  diameter, 
at  a  distance  AD  from  it, 

equal  to  the  side  of  the     i\.  T?B 

given  square  C ;  from  the  point  E,  where  the  parallel  cuts  the 
circumference,  draw  EF  perpendicular  to  the  diameter ;  AF 
and  FB  will  be  the  sides  of  the  rectangle  required. 

For  their  sum  is  equal  to  AB  ;  and  their  rectangle  AF.FB  is 
equivalent  to  the  square  of  EF,  or  to  the  square  of  AD ;  hence 
that  rectangle  is  equivalent  to  the  given  square  C. 

Scholium.  To  render  the  problem  possible,  the  distance  AD 
must  not  exceed  the  radius ;  that  is,  the  side  of  the  square  C 
must  not  exceed  the  half  of  the  line  AB. 


BOOK  IV. 


107 


PROBLEM  XVII. 


To  construct  a  rectangle  that  shall  he  equivalent  to  a  given 
square,  and  the  difference  of  whose  adjacent  sides  shall  he 
equal  to  a  given  line. 

Suppose  C  equal  to  the  given  square,  and  AB  the  difference 
of  the  sides. 

Upon  the  given  line  AB  as  a  diame- 
ter, describe  a  semicircle  :  at  the  ex- 
tremity of  the  diameter  draw  the  tan- 
gent AD,  equal  to  the  side  of  the  square 
C  ;  through  the  point  D  and  the  centre 
O  draw  the  secant  DF ;  then  will  DE 
and  DF  be  the  adjacent  sides  of  the 
rectangle  required. 

For,  first,  the  difference  of  these  sides 
is  equal  to  the  diameter  EF  or  AB  ; 
secondly,  the  rectangle   DE,  DF,   is 
equal  to  AD^  (Prop.  XXX.) ;  hence  that  rectangle  is  equivalent 
to  the  given  square  C. 


PROBLEM  XVIII. 


To  find  the  common  measure,  if  there  is  one,  between  the  diagonal 
and  the  side  of  a  square. 


Let  ABCG  be  any  square  what- 
ever, and  AC  its  diagonal. 

We  must  first  apply  CB  upon 
CA,  as  often  as  it  may  be  contained 
there.  For  this  purpose,  let  the 
semicircle  DBE  be  described,  from 
the  centre  C,  with  the  radius  CB. 
It  is  evident  that  CB  is  contained 
once  in  AC,  with  the  remainder 
AD ;  the  result  of  the  first  operation 
is  therefore  the  quotient  1,  with  the  remainder  AD,  which  lat- 
ter must  now  be  compared  with  BC,  or  its  equal  AB. 

We  might  here  take  AF=AD,  and  actually  apply  it  upon 
AB  ;  we  should  find  it  to  be  contained  twice  with  a  remain- 
der :  but  as  that  remainder,  and  those  which  succeed  it,  con- 


108 


GEOMETRY. 


linue  diminishing,  and  would  soon 
elude  our  comparisons  by  their  mi- 
nuteness, this  would  be  but  an  imper- 
fect mechanical  method,  from  which 
no  conclusion  could  be  obtained  to 
determine  whether  the  lines  AC,  CB, 
have  or  have  not  a  common  measure. 
There  is  a  very  simple  way,  however, 
of  avoiding  these  decreasing  lines, 
and  obtaining  the  result,  by  operating 
only  upon  lines  which  remain  always  of  the  same  magnitude. 

The  angle  ABC  being  a  right  angle,  AB  is  a  tangent,  and 
AE  a  secant  drawn  from  the  same  point ;  so  that  AD  :  AB  :  : 
AB  :  AE  (Prop.  XXX.).  Hence  in  the  second  operation,  when 
AD  is  compared  with  AB,  the  ratio  of  AB  to  AE  may  be  taken 
instead  of  that  of  AD  to  AB ;  now  AB,  or  its  equal  CD,  is  con- 
tained twice  in  AE,  with  the  remainder  AD  ;  the  result  of  the 
second  operation  is  therefore  the  quotient  2  with  the  remain- 
der AD,  which  must  be  compared  with  AB. 

Thus  the  third  operation  again  consists  in  comparing  AD 
with  AB,  and  may  be  reduced  in  the  same  manner  to  the  com- 
parison of  AB  or  its  equal  CD  with  AE  ;  from  which  there  will 
again  be  obtained  2  for  the  quotient,  and  AD  for  the  re- 
mainder. 

Hence,  it  is  evident  that  the  process  will  never  terminate  ; 
and  therefore  there  is  no  common  measure  between  the  diago- 
nal and  the  side  of  a  square  :  a  truth  which  was  already  known 
by  arithmetic,  since  these  two  lines  are  to  each  other  :  :  \/2  :  1 
(Prop.  XI.  Cor.  4.),  but  which  acquires  a  greater  degree  of 
clearness  by  the  geometrical  investigation. 


BOOK  V.  109 


BOOK  V. 


REGULAR  POLYGONS,  AND  THE  MEASUREMENT  OF  THE 
CIRCLE. 


Befinitioru 

A  Polygon,  which  is  at  once  equilateral  and  equiangular,  is 
called  a  regular  polygon. 

Regular  polygons  may  have  any  number  of  sides :  the  equi- 
lateral triangle  is  one  of  three  sides ;  the  square  is  one  of  Jour. 


PROPOSITION  I.    THEOREM. 

Two  regular  polygons  of  the  same  number  of  sides  are  similar 

figures. 

Suppose,  for  example, 
that  ABCDEF,  ahcdef 
are  two  regular;  hexa- 
gons. The  sum  of  all  the 
angles  is  the  same  in  both 
figures,being  in  each  equal 
to  eight  right  angles  (Book  I.  Prop.  XXVI.  Cor.  3.).  The  angle 
A  is  the  sixth  part  of  that  sum ;  so  is  the  angle  a  :  hence  the 
angles  A  and  a  are  equal ;  and  for  the  same  reason,  the  angles 
B  and  ft,  the  angles  C  and  c,  &c.  are  equal. 

Again,  since  the  polygons  are  regular,  the  sides  AB,  BC,  CD, 
&c.  are  equal,  and  likewise  the  sides  ah,  he,  cd,  &c.  (Def.)  ;  it  is 
plain  that  AB  :  ah  :  :  BC  :  he  :  :  CD  :  cd,  &c. ;  hence  the 
two  figures  in  question  have  their  angles  equal,  and  their  ho- 
mologous sides  proportional ;  consequently  they  are  similar 
(Book  IV.  Def.  1.). 

Cor.  The  perimeters  of  two  regular  polygons  of  the  same 
number  of  sides,  are  to  each  other  as  their  homologous  sides, 
and  their  surfaces  are  to  each  other  as  the  squares  of  those  sides 
(Book  IV.  Prop.  XXVII.). 

Scholium.  The  angle  of  a  regular  polygon,  like  the  angle  of 
an  equiangular  polygon,  is  determined  by  the  number  of  its 
sides  (Book  I.  Prop.  XXVI.). 

K 


no  GEOMETRY.  % 


PROPOSITION  II.    THEOREM. 

Any  regular  polygon  may  he  inscribed  in  a  circle^  and  circum- 
scribed about  one. 

Let  ABCDE  &c.  be  a  regular  poly- 
gon :  describe  a  circle  through  the  three 
points  A,  B,  C,  the  centre  being  O,  and 
OP  the  perpendicular  let  fall  from  it,  to 
the  middle  point  of  EC  :  draw  AO  and 
OD. 

If  the  quadrilateral  OPCD  be  placed 
upon  the  quadrilateral  OPBA,  they  will 
coincide  ;  for  the  side  OP  is  common  ; 
the  angle  OPC=OPB,  each  being  a  right  angle  ;  hence  the 
side  PC  will  apply  to  its  equal  PB,  and  the  point  C  will  fall  on 
B :  besides,  from  the  nature  of  the  polygon,  the  angle  PCD  = 
PBA ;  hence  CD  will  take  the  direction  BA ;  and  since  CD  = 
BA,  the  point  D  will  fall  on  A,  and  the  two  quadrilaterals  will 
entirely  coincide.  The  distance  OD  is  therefore  equal  to  AO  ; 
and  consequently  the  circle  which  passes  through  the  three 
points  A,  B,  C,  will  also  pass  through  the  point  D.  By  the 
same  mode  of  reasoning,  it  might  be  shown,  that  the  circle 
which  passes  through  the  three  points  B,  C,.D,  will  also  pass 
through  the  point  E  ;  and  so  of  all  the  rest :  hence  the  circle 
which  passes  through  the  points  A,  B,  C,  passes  also  through 
the  vertices  of  all  the  angles  in  the  polygon,  which  is  therefore 
inscribed  in  this  circle. 

Again,  in  reference  to  this  circle,  all  the  sides  AB,  BC,  CD, 
&c.  are  equal  chords ;  they  are  therefore  equally  distant  from 
the  centre  (Book  III.  Prop.  VIII.) :  hence,  if  from  the  point  O 
with  the  distance  OP,  a  circle  be  described,  it  will  touch  the 
side  BC,  and  all  the  other  sides  of  the  polygon,  each  in  its  mid- 
dle point,  and  the  circle  will  be  inscribed  in  the  polygon,  or  the 
polygon  described  about  the  circle. 

Scholium  1.  The  point  O,  the  common  centre  of  the  in- 
scribed and  circumscribed  circles,  may  also  be  regarded  as  the 
centre  of  the  polygon ;  and  upon  this  principle  the  angle  AOB 
is  called  the  angle  at  the  centre^  being  formed  by  two  radii 
drawn  to  the  extremities  of  the  same  side  AB. 

Since  all  the  chords  AB,  BC,  CD,  &c.  are  equal,  all  the  an- 
gles at  the  centre  must  evidently  be  equal  likewise  ;  and  there- 
fore the  value  of  each  will  be  found  by  dividing  four  right  an- 
gles by  the  number  of  sides  of  the  polygon. 


BOOK  V. 


Ill 


Scholium  2.  To  inscribe  a  regu- 
lar polygon  of  a  certain  number  of 
sides  in  a  given  circle,  we  have  only 
to  divide  the  circumference  into  as 
many  equal  parts  as  the  polygon 
has  sides :  for  the  arcs  being  equal, 
the  chords  AB,  BC,  CD,  &c.  will 
also  be  equal ;  hence  likewise  the 
triangles  AOB,  BOC,  COD,  must 
be  equal,  because  the  sides  are  equal  each  to  each  ;  hence  all 
the  angles  ABC,  BCD,  CDE,  6lc,  will  be  equal ;  hence  the 
figure  ABCDEH,  will  be  a  regular  polygon. 


PROPOSITION  III.      PROBLEM. 


To  inscribe  a  square  in  a  given  circle. 


Draw  two  diameters  AC,  BD,  cut- 
ting each  other  at  right  angles  ;  join 
their  extremities  A,  B,  C,  D :  the  figure 
ABCD  will  be  a  square.  For  the  an- 
gles AOB,  BOC,  &c.  being  equal,  the 
chords  AB,  BC,  &c.  are  also  equal : 
and  the  angles  ABC,  BCD,  &c.  being 
in  semicircles,  are  right  angles. 


Scholium.  Since  the  triangle  BCO  is  right  angled  and  isos- 
celes, we  have  BC  :  BO  :  :  V2  :  1  (Book  IV.  Prop.  XL 
Cor.  4.) ;  hence  the  side  of  the  inscribed  square  is  to  the  radius^ 
as  the  square  root  of  2,  is  to  unity. 


PROPOSITION  IV.      PROBLEM. 

In  a  given  circkf  to  inscribe  a  regular  hexagon  and  an  equilate- 
ral triangle. 


i 


112 


GEOMETRY. 


Suppose  the  problem  solved, 
and  that  AB  is  a  side  of  the  in- 
scribed hexagon ;  the  radii  AO, 
OB  being  drawn,  the  triangle 
AOB  will  be  equilateral. 

For,  the  angle  AOB  is  the  sixth 
part  of  four  right  angles  ;  there- 
fore, taking  the  right  angle  for 
unity,  we  shall  have  AOB=|r: 
f :  and  the  two  other  angles 
ABO,  BAO,  of  the  same  trian- 
gle, are  together  equal  to  2 — | 
=f  ;  and  being  mutually  equal, 
each  of  them  must  be  equal  to  | ;  hence  the  triangle  ABO  is 
equilateral ;  therefore  the  side  of  the  inscribed  hexagon  is  equal 
to  the  radius. 

Hence  to  inscribe  a  regular  hexagon  in  a  given  circle,  the 
radius  must  be  applied  six  times  to  the  circumference  ;  which 
will  bring  us  round  to  the  point  of  beginning. 

And  the  hexagon  ABCDEF  being  inscribed,  the  equilateral 
triangle  ACE  may  be  formed  by  joining  the  vertices  of  the 
alternate  angles. 

Scholium,  The  figure  ABCO  is  a  parallelogram  and  even 
a  rhombus,  since  AB=BC=CO=AO  ;  hence  the  sum  of  the 
SCJUCires  of  the  diagonals  AC^+BO^  is  equivalent  to  the  sum  of 
the  squares  of  the  sides,  that  is,  to  4AB^  or  4B0^  (Book  IV. 
Prop  XIV.  Cor.) :  and  taking  away  BO^  from  both,  there  will 
remain  AC2=3B02;  hence  AC2  :  BO^  :  :  3  :  l,or  AC  :  BO 
:  :  V3  :  1 ;  hence  the  side  of  the  inscribed  equilateral  triangle 
is  to  the  radius  as  the  square  root  of  three  is  to  unity. 


PROPOSITION  V.      PROBLEM. 

In  a  given  circle^  to  inscribe  a  regular  decagon^  then  apentagon^ 
and  also  a  regular  polygon  of  fifteen  sides. 


♦»« 


BOOK  V. 


113 


Divide  the  radius  AO  in 
extreme  and  mean  ratio  at 
the  point  M  (Book  IV.  Prob. 
IV.) ;  take  the  chord  AB  equal 
to  OM  the  greater  segment ; 
AB  will  be  the  side  of  the 
regular  decagon,  and  will  re- 
quire to  be  applied  ten  times 
to  the  circumference. 

For,  drawing  MB,  we  have 
by  construction,  AO  :  OM 
:  :  OM  :  AM  ;  or,  since  AB 
=0M,  AO  :  AB  :  :  AB  : 
AM  ;  since  the  triangles  ABO,  AMB,  have  a  common  angle  A, 
included  between  proportional  sides,  they  are  similar  (Book 
IV.  Prop.  XX.).  Now  the  fxiangle  OAB  being  isosceles,  AMB 
must  be  isosceles  also,  and  AB==BM  ;  but  AB=OM  ;  hence 
also  MB  r=  OM  ;  hence  the  triangle  BMO  is  isosceles. 

Again,  the  angle  AMB  being  exterior  to  the  isosceles  trian- 
gle BMO,  is  double  of  the  interior  angle  O  (Book  I.  Prop. 
XXV.  Cor.  6.)  :  but  the  angle  AMB=MAB  ;  hence  the  trian- 
gle OAB  is  such,  that  each  of  the  angles  OAB  or  OBA,  at  its 
base,  is  double  of  O,  the  angle  at  its  vertex  ;  hence  the  three 
angles  of  the  triangle  are  together  equal  to  five  times  the  angle 
O,  which  consequently  is  the  fifth  part  of  the  two  right  angles, 
or  the  tenth  part  of  four  ;  hence  the  arc  AB  is  the  tenth  part 
of  the  circumference,  and  the  chord  AB  is  the  side  of  the  reg- 
ular decagon. 

2d.  By  joining  the  alternate  corners  of  the  regular  decagon, 
the  pentagon  ACEGI  will  be  formed,  also  regular. 

3d.  AB  being  still  the  side  of  the  decagon,  let  AL  be  the 
side  of  a  hexagon  ;  the  arc  BL  will  then,  with  reference  to 
the  whole  circumference,  be  } — yV,  or  yV  ;  hence  the  chord  BI^ 
will  be  the  side  of  the  regular  polygon  of  fifteen  sides,  or  pente- 
decagon.  It  is  evident  also,  that  the  arc  CL  is  the  third  of  CB. 

Scholium.  Any  regular  polygon  being  inscribed,  if  the  arcs 
subtended  by  its  sides  be  severally  bisected,  the  chords  of  those 
semi-arcs  will  form  a  new  regular  polygon  of  double  the  num- 
ber of  sides  :  thus  it  is  plain,  that  the  square  will  enable  us  to  in- 
scribe successively  regular  polygons  of  8, 16, 32,  &c.  sides.  And 
m  like  manner,  by  means  of  the  hexagon,  regular  polygons  of 
12,  24,  48,  i&c.  sides  may  be  inscribed  ;  by  means  of  the  deca- 
gon, polygons  of  20, 40, 80,  &lc.  sides  ;  by  means  of  the  pente- 
decagon,  polygons  of  30,  60, 120,  &c.  sides. 

It  is  further  evident,  that  any  of  the  inscribed  polygons  will 
be  less  than  the  inscribed  polygon  of  double  the  number  of 
fiides,  since  a  part  is  less  than  the  whole. 
K*    15 


114 


GEOMETRY. 


ii^v;; 


PROPOSITION  VI.    PROBLEM. 

A  regular  inscHhed  polygon  being  given,  to  circumscribe  a  sim- 
ilar polygon  about  the  same  circle. 

Let  CBAFED  be  a  regular  polygon. 

At  T,  the  middle  point 
of  the  arc  AB,  apply  the 
tangent  GH,  which  will 
be  parallel  to  AB  (Book 
III.  Prop.  X.)  ;  do  the 
same  at  the  middle  point 
of  each  of  the  arcs  BC, 
CD,  &c. ;  these  tangents, 
by  their  intersections, 
will  form  the  regular 
circumscribed  polygon 
GHIK  &c.  similar  to 
the  one  inscribed.  

k:       Q         t 

Since  T  is  the  middle  point  of  the  arc  BTA,  and  N  the  mid- 
dle point  of  the  equal  arc  BNC,  it  follows,  that  BT=BN  ;  or 
that  the  vertex  B  of  the  inscribed  polygon,  is  at  the  middle 
point  of  the  arc  NBT.  Draw  OH.  The  line  OH  will  pass 
through  the  point  B. 

For,  the  right  angled  triangles  OTH,  OHN,  having  the  com- 
mon hypothenuse  OH,  and  the  side  OT=:ON,  must  be  equal 
(Book  I.  Prop.  XVn.),  and  consequently  the  angle  TOHrr: 
HON,  wherefore  the  line  OH  passes  through  the  middle  point 
B  of  the  arc  TN.  For  a  like  reason,  the  point  I  is  in  the  pro- 
longation of  OC  ;  and  so  with  the  rest. 

But,  since  GH  is  parallel  to  AB,  and  HI  to  BC,  the  angle 
GHI=ABC  (Book  I.  Prop.  XXIV.) ;  in  like  manner  HIK=: 
BCD  ;  and  so  with  all  the  rest :  hence  the  angles  of  the  cir- 
cumscribed polygon  are  equal  to  those  of  the  inscribed  one. 
And  further,  by  reason  of  these  same  parallels,  we  have  GH  : 
AB  :  :  OH  :  OB,  and  HI  :  BC  :  :  OH  :  OB ;  therefore  GH  : 
AB  :  :  HI  :  BC.  But  AB=BC,  therefore  GH=HI.  For  the 
same  reason,  HI = IK,  &c.;  hence  the  sides  of  the  circum- 
scribed polygon  are  all  equal ;  hence  this  polygon  is  regular, 
and  similar  to  the  inscribed  one. 


Cor.  1.  Reciprocally,  if  the  circumscribed  polygon  GHIK 
&c.  were  given,  and  the  inscribed  one  ABC  &c.  were  re- 
quired to  be  deduced  from  it,  it  would  only  be  necessary  to 


BOOK  V. 


115 


draw  from  the  angles  G,  H,  I,  &c 
lines  OG,  OH,  &c.  meeting  the 
A,  B,  C,  &c. ;  then  to  join 
AB,  BC,  &c. ;  this  would  form 
easier  solution  of  this  problem 
points  of  contact  T,  N,  P,  &c. 
which  likewise  would  form  an 
the  circumscribed  one. 


.  of  the  given  polygon,  straight 
circumference  in  the  points 

those  points  by  the  chords 
the  inscribed  polygon.     An 

would  be  simply  to  join  the 

by  the  chords  TN,  NP,  &c. 

inscribed  polygon  similar  to 


Cor.  2.  Hence  we  may  circumscribe  about  a  circle  any 
regular  polygon,  which  can  be  inscribed  within  it,  and  con- 
versely. 

Cor,  3.     It  is  plain  that  NH  +  HT~HT  +  TG=HG,  one  of 

the  equal  sides  of  the  polygon. 


PROPOSITION  VII.    PROBLEM. 


A  circle  and  regular  circumscribed  polygon  being  given,  it  is 
required  to  circumscribe  the  circle  by  another  regular  polygon 
having  double  the  number  of  sides. 


i     ^     /v 


E 


Let  the  circle  whose  centre  is  P,  be  circumscribed  by  the 
square  CDEG  :  it  is  required  to  find  a  regular  circumscribed 
octagon. 

Bisect  the  arcs  AH,  HB,  BF, 
FA,  and  through  the  middle 
points  c,  dy  a,  b,  draw  tangents  to 
the  circle,  and  produce  them  till 
they  meet  the  sides  of  the  square : 
then  will  the  figure  ApHdB  &c, 
be  a  regular  octagon. 

For,  having  drawn  P<i,  P<z,  let 
the  quadrilateral  VdgB,  be  ap- 
plied to  the  quadrilateral  PB/a, 
so  that  PB  shall  fall  on  PB. 
Then,  since  the   angle  dZPB 


/ 

-^ 

> 

( 

/\ 

P 

\     , 

P 

\ 

\} 

\ 

.^ 

J 

c 


7t    ^    k 


smce  tne  angle  arJts  is 
equal  to  the  angle  BP<2,  each  being  half  a  right  angle,  the  line 
Yd  will  fall  on  its  equal  P«,  and  the  point  d  on  the  point  a.  But 
the  angles  Vdg,  Faf,  are  right  angles  (Book  HI.  Prop.  IX.)  ; 
hence  the  line  dg  will  take  the  direction  of.  The  angles  PB^, 
PB/,  are  also  right  angles  ;  hence  B^  will  take  the  direction 
B/*;  therefore,  the  two  quadrilaterals  will  coincide,  and  the 
point  g  will  fall  at/;  hence,  B^=B/,  dg=af,  and  the  angle 
dgB=Bfa.  By  applying  in  a  similar  manner,  the  quadrilate- 
rals PBfa,  PF/i<z,  it  may  be  shown,  that  afz^ah,  /B=F/i,  and 
the  angle  B/arraAF.    But  since  the  two  tangents /r,  /B^  are 


116  GEOMETRY. 

equal  (Book  III.  Prob.  XIV.  Sch.),  it  follows  that  fh,  which  is 
twice /a,  is  equal  to^,  which  is  twice /B. 

In  a  similar  manner  it  may  be  shown  that  hfznhi,  and  the 
angle  Fz^=FAa,  or  that  any  two  sides  or  any  two  angles  of  the 
octagon  are  equal :  hence  the  octagon  is  a  regular  polygon  (Def ). 
The  construction  which  has  been  made  in  the  case  of  the  square 
and  the  octagon,  is  equally  applicable  to  other  polygons. 

,  Cor,  It  is  evident  that  the  circumscribed  square  is  greater  than 
the  circumscribed  octagon  by  the  four  triangles,  Cnp,  kJyg, 
KEf,  Git ;  and  if  a  regular  polygon  of  sixteen  sides  be  circum- 
scribed about  the  circle,  we  may  prove  in  a  similar  way,  that 
the  figure  having  the  greatest  number  of  sides  will  be  the  least ; 
and  the  same  may  be  shown,  whatever  be  the  number  of  sides 
of  the  polygons  :  hence,  in  general,  ant/  circumscribed  regular 
polygon,  will  be  greater  than  a  circumscribed  regular  polygon 
having  double  the  number  of  sides. 


PROPOSITION  VIII.     THEOREM. 

Two  regular  polygons,  of  the  same  number  of  sides,  can  always 
be  formed,  the  one  circumscribed  about  a  circle,  the  other  in- 
scribed in  it,  which  shall  differ  from  each  other  by  less  than 
any  assignable  surface. 

Let  Q  be  the  side  of  a  square 
less  than  the  given  surface. 
Bisect  AC,  a  fourth  part  of 
the  circumference,  and  then  bi 
sect  the  half  of  this  fourth,  and 
proceed  in  this  manner,  always 
bisecting  one  of  the  arcs  formed 
by  the  last  bisection,  until  an 
arc  is  found  whose  chord  AB  is 
less  than  Q.  As  this  arc  will 
be  an  exact  part  of  the  circum- 
ference, if  we  apply  chords  AB, 
BC,  CD,  &c.  each  equal  to  AB,  the  last  will  terminate  at  A, 
and  there  will  be  formed  a  regular  polygon  ABCDE  &c.  in 
the  circle. 

Next,  describe  about  the  circle  a  similar  polygon  abcde  &c. 
(Prop.  VI.) :  the  difference  of  these  two  polygons  will  be  less 
than  the  square  of  Q. 

For,  from  the  points  a  and  b,  draw  the  lines  aO,  bO,  to  the 
centre  O  :  they  will  pass  through  the  points  A  and  B,  as  was 


BOOK  V.  117 

shown  in  Prop.  VI.  Draw  also  OK  to  the  point  of  contact 
K  ;  it  will  bisect  AB  in  I,  and  be  perpendicular  to  it  (Book  III. 
Prop.  VI.  Sch.).     Produce  AO  to  E,  and  draw  BE. 

Let  P  represent  the  circumscribed  polygon,  and  p  the  in- 
scribed polygon  :  then,  since  the  triangles  aOh,  AOB,  are  like 
parts  of  P  and  p,  we  shall  have 

aOh  :  AOB  :  :  P  :  jo  (Book  II.  Prop.  XI.) : 
But  the  triangles  being  similar, 

aOh  :  AOB  :  :  Oa"  :  OA^,  or  OK^. 
Hence,  P  :  ^  ;  :  Oa^  :  OKI 

Again,  since  the  triangles  OaK,  EAB  are  similar,  having 
their  sides  respectively  parallel, 

Oa"  :  0K2  :  :  AE^  :  EB^,    hence, 
F  :  p  :  :  AE^  :  EB^,  or  by  division, 

P  ;  P— j9  :  :  AE^  :  AE^— EB^,  or  AB^. 
But  P  is  less  than  the  square  described  on  the  diameter  AE 
(Prop.  VII.  Cor.)  ;  therefore  P — p  is  less  than  the  square  de- 
scribed on  AB  ;  that  is,  less  than  the  given  square  on  Q :  hence 
the  difference  between  the  circumscribed  and  inscribed  poly- 
gons may  always  be  made  less  than  a  given  surface. 

Cor.  1.  A  circumscribed  regular  polygon,  having  a  given 
number  of  sides,  is  greater  than  the  circle,  because  the  circle 
makes  up  but  a  part  of  the  polygon  :  and  for  a  like  reason,  the 
inscribed  polygon  is  less  than  the  circle.  But  by  increasing 
the  number  of  sides  of  the  circumscribed  polygon,  the  polygon 
is  diminished  (Prop.  VII.  Cor.),  and  therefore  approaches  to 
an  equality  with  the  circle  ;  and  as  the  number  of  sides  of  the 
inscribed  polygon  is  increased,  the  polygon  is  increased  (Prop. 
V.  Sch.),  and  therefore  approaches  to  an  equality  with  the 
circle. 

Now,  if  the  number  of  sides  of  the  polygons  be  indefinitely  in- 
creased, the  length  of  each  side  will  be  indefinitely  small,  and  the 
polygons  will  ultimately  become  equal  to  each  other,  and  equal 
also  to  the  circle. 

For,  if  they  are  not  ultimately  equal,  let  D  represent  their 
smallest  difference. 

Now,  it  has  been  proved  in  the  proposition,  that  the  diffe- 
rence between  the  circumscribed  and  inscribed  polygons,  can 
be  made  less  than  any  assignable  quantity :  that  is,  less  than 
D  :  hence  the  difference  between  the  polygons  is  equal  to  D, 
and  less  than  D  at  the  same  time,  which  is  absurd  :  therefore, 
the  polygons  are  ultimately  equal.  But  when  they  are  equal 
to  each  other,  each  must  also  be  equal  to  the  circle,  since  the 
circumscribed  polygon  cannot  fall  within  the  circle,  nor  the 
inscribed  polygon  without  it. 


118        "^  GEOMETRY. 

Cor.  2,  Since  the  circumscribed  polygon  has  the  same  num- 
ber of  sides  as  the  corresponding  inscribed  polygon,  and  since 
the  two  polygons  are  regular,  they  will  be  similar  (Prop.  I.) ; 
and  therefore  when  they  become  equal,  they  will  exactly  coin- 
cide, and  have  a  common  perimeter.  But  as  the  sides  of  the 
circumscribed  polygon  cannot  fall  within  the  circle,  nor  the 
sides  of  the  inscribed  polygon  without  it,  it  follows  that  the 
perimeters  of  the  polygons  will  unite  on  the  circumference  of  the 
circle,  and  become  equal  to  it. 

Cor.  3.  When  the  number  of  sides  of  the  inscribed  polygon 
is  indefinitely  increased,  and  the  polygon  coincides  with  the 
circle,  the  line  01,  drawn  from  the  centre  O,  perpendicular  to 
the  side  of  the  polygon,  will  become  a  radius  of  the  circle,  and 
any  portion  of  the  polygon,  as  ABCO,  will  become  the  sector 
OAKBC,  and  the  part  of  the  perimeter  AB  +  BC,  will  become 
the  arc  AKBC. 


PROPOSITIOJM  IX.     THEOREM. 

The  area  of  a  regular  polygon  is  equal  to  its  perimeter,  multi- 
plied by  haf  the  radius  of  the  inscribed  circle. 

Let  there  be  the  regular  polygon 
GHIK,  and  ON,  OT,  radii  of  the  in- 
scribed circle.  The  triangle  GOH 
will  be  measured  by  GH  x  iOT ;  the 
triangle  OHI,  by  HIx|6N:  but 
ON=OT;  hence  the  two  triangles 
taken  together  will  be  measured  by 
(GH  +  HI)xiOT.  And,  by  con- 
tinuing the  same  operation  for  the 
other  triangles,  it  will  appear  that 
the  sum  of  them  all,  or  the  whole 
polygon,  is  measured  by  the  sum  of  the  bases  GH,  HI,  &c. 
or  the  perimeter  of  the  polygon,  multiplied  into  |0T,  or  half 
the  radius  of  the  inscribed  circle. 

Scholium.  The  radius  OT  of  the  inscribed  circle  is  nothing 
else  than  the  perpendicular  let  fall  from  the  centre  on  one  of 
the  sides ;  it  is  sometimes  named  the  apothem  of  the  polygon. 


BOOK  V. 


119 


PROPOSITION  X.    THEOREM. 


The  perimeters  of  two  regular  polygons,  having  the  same  num- 
ber of  sides,  are  to  each  other  as  the  radii  of  the  circumscribed 
circles,  and  also,  as  the  radii  of  the  inscribed  circles  ;  and  their 
areas  are  to  each  other  as  the  squares  of  those  radii. 

Let  AB  be  the  side  of  the  one  poly- 
gon, O  the  centre,  and  consequently 
OA  the  radius  of  the  circumscribed 
circle,  and  CD,  perpendicular  to  AB, 
the  radius  of  the  inscribed  circle;  let 
ab,  in  like  manner,  be  a  side  of  the 
other  polygon,  o  its  centre,  oa  and  od 
the  radii  of  the  circumscribed  and  the 
inscribed  circles.  The  perimeters  of 
the  two  polygons  are  to  each  other  as  the  sides  AB  and  ah 
(Book  IV.  Prop.  XXVII.) :  but  the  angles  A  and  a  are  equal, 
being  each  half  of  the  angle  of  the  polygon  ;  so  also  are  the 
angles  B  and  b ;  hence  the  triangles  ABO,  abo  are  similar,  as 
are  likewise  the  right  angled  triangles  ADO,  ado-,  hence 
AB  '.  ab  :  :  AO  :  ao  :  :  DO  :  do  ;  hence  the  perimeters  of  the 
polygons  are  to  each  other  as  the  radii  AO,  ao  of  the  circum- 
scribed circles,  and  also,  as  the  radii  DO,  do  of  the  inscribed 
circles. 

The  surfaces  of  these  polygons  are  to  each  other  as  the 
squares  of  the  homologous  sides  AB,  ab  ;  they  are  therefore 
likewise  to  each  other  as  the  squares  of  AO,«o,the  radii  of  the 
circumscribed  circles,  or  as  the  squares  of  OD,  od^  the  radii  of 
the  inscribed  circles. 


PROPOSITION  XI.     THEOREM. 


The  circumferences  of  circles  are  to  each  other  as  their  radii, 
and  the  areas  are  to  each  other  as  the  squares  of  their  radii. 


120 


GEOMETRY. 


.  J, 


Let  us  designate  the  circumference  of  the  circle  whose  radius 
is  CA  by  drc,  CA ;  and  its  area,  by  area  CA :  it  is  then  to  be 
shown  that 

circ.  CA  :  circ,  OB  :  :  CA  :  OB,  and  that 
area  CA  :  area  OB  :  :  C A^  :  OB^. 


Inscribe  within  the  circles  two  regular  polygons  of  the  same 
number  of  sides.  Then,  whatever  be  the  number  of  sides, 
their  perimeters  will  be  to  each  other  as  the  radii  CA  and  OB 
(Prop.  X.).  Now,  if  the  arcs  subtending  the  sides  of  the  poly- 
gons be  continually  bisected,  until  the  number  of  sides  of  the 
polygons  shall  be  indefinitely  increased,  the  perimeters  of  the 
polygons  will  become  equal  to  the  circumferences  of  the  cir- 
cumscribed circles  (Prop.  VIII.  Cor.  2.),  and  we  shall  have 
circ.  CA  :  circ.  OB  :  :  CA  :  OB. 

Again,  the  areas  of  the  inscribed  polygons  are  to  each  other 
as  CA^  to  OB^  (Prop.  X.).  But  when  the  number  of  sides  of 
the  polygons  is  indefinitely  increased,  the  areas  of  the  polygons 
become  equal  to  the  areas  of  the  circles,  each  to  each,  (Prop. 
VIII.  Cor.  1.)  ;  hence  we  shall  have 

area  CA  :  area  OB  :  :  CA^  :  OB^. 


V 


Cor.  The  similar  arcs  AB, 
DE  are  to  each  other  as  their 
radii  AC,  DO  ;  and  the  similar 
sectors  ACB,  DOE,  are  to  each 
other  as  the  squares  of  their 
radii. 

For,  since  the.  arcs  are  simi-  ^ 

lar,  the  angle  C  is  equal  to  the  angle  O  (Book  IV.  Def.  3.)  ; 
but  C  is  to  four  right  angles,  as  the  arc  AB  is  to  the  whole  cir- 
cumference described  with  the  radius  AC  (Book  III.  Prop. 
XVII.) ;  and  O  is  to  the  four  right  angles,  as  the  arc  DE  is  to 
the  circumference  described  with  the  radius  OD  :  hence  the 
arcs  AB,  DE,  are  to  each  other  as  the  circumferences  of  which 


BOOK  V. 


121 


they  form  part :  but  these  circumferences  are  to, each  other  as 
their  radii  AC,  DO  ;  hence 

arc  AB  :  arci)E  :  :  AC  :  DO. 
For  a  Hke  reason,  the  sectors  ACB,  DOE  are  to  each  other 
as  the  whole  circles  ;  which  again  are  as  the  squares  of  their 
radii ;  therefore 

sect.  ACB  :  sect.BOJE.  :  :  AC^  :  DO2. 


PROPOSITION  XII.     THEOREM. 

The  area  of  a  circle  is  equal  to  the  product  of  its  circumference  by 
half  the  radius. 


Let  ACDE  be  a  circle  whose 
centre  is  O  and  radius  OA  :  then 
will 

area  OA=|OA  x  circ.  OA. 

For,  inscribe  in  the  circle  any 
regular  polygon,  and  draw  OF 
perpendicular  to  one  of  its  sides. 
Then  the  area  of  the .  polygon 
will  be  equal  to  ^OF,  multiplied 
by  the  perimeter  (Prop.  IX.). 
Now,  let  the  number  of  sides  of  the  polygon  be  indefinitely 
increased  by  continually  bisecting  the  arcs  which  subtend  the 
sides.:  the  perimeter  will  then  become  equal  to  the  circumfe- 
rence of  the  circle,  the  perpendicular  OF  will  become  equal  to 
OA,  and  the  area  of  the  polygon  to  the  area  of  the  circle 
(Prop.  VIII.  Cor.  1.  &  3.).  But  the  expression  for  the  area 
will  then  become  • 

'  area  0A=^0  Ax  circ.  OA: 
consequently,  the  area  of  a  circle  is  equal  to  the  product  of 
half  the  radius  of  the  circumference. 


Cor.  1.  The  area  of  a  sector  is  equal 
to  the  arc  of  that  sector  multiplied  by  half 
its  radius. 

For,  the  sector  ACB  is  to  the  whole 
circle  as  the  arc  AMB  is  to  the  whole 
circumference  ABD  (Book  III.  Prop. 
XVII.  Sch.  2.),  or  as  AMBx^AC  is  to 
ABDx^AC.  But  the  whole  circle  is 
equal  to  ABD  x  ^AC  ;  hence  the  sector 
ACB  is  measured  by  AMB  x  i  AC. 

L   16 


122J  GEOMETRY. 

Cor.  2.  Let  the  circumference  of  the 
circle  whose  diameter  is  unity,  be  denoted 
by  7t :  then,  because  circumferences  are 
to  each  other  as  their  radii  or  diameters, 
we  shall  have  the  diameter  1  to  its  cir- 
cumference 7t,  as  the  diameter  2CA  is 
to  the  circumference  whose  radius  is  CA, 
that  is,  1  :  7t  :  :  2CA  :  circ.  CA,  there- 
fore czrc.  CA=7rx2CA.  Multiply  both 
terms  by  ^CA  ;  we  have  iCA  x  circ.  CA 
=7tx  C\\  or  area  CA=7tx  CA^  ;  hence  the  area  of  a  circle  is 
equal  to  the  product  of  the  square  of  its  radius  by  the  constant 
number  tt,  which  represents  the  circumference  whose  diameter 
is  1,  or  the  ratio  of  the  circumference  to  the  diameter. 

In  like  manner,  the  area  of  the  circle,  whose  radius  is  OB, 
will  be  equal  to  ^  x  OB^ ;  but  tt  x  CA^  :  tt  x  OB^  :  :  CA^  :  OB^  ; 
hence  the  areas  of  circles  are  to  each  other  as  the  squares  of 
their  radii,  which  agrees  with  the  preceding  theorem. 

Scholium.  We  have  already  observed,  that  the  problem  of 
the  quadrature  of  the  circle  consists^  in  finding  a  square  equal 
in  surface  to  a  circle,  the  radius  of  which  is  known.  Now  it 
has  just  been  proved,  that  a  circle  is  equivalent  to  the  rectangle 
contained  by  its  circumference  and  half  its  radius  ;  and  this 
rectangle  may  be  changed  into  a  square,  by  finding  a  mean 
proportional  between  its  length  and  its  breadth  (Book  IV. 
Prob.  III.).  To  square  the  circle,  therefore,  is  to  find  the  cir- 
cumference when  the  radius  is  given  ;  and  for  effecting  this,  it 
is  enough  to  know  the  ratio  of  the  circumference  to  its  radius, 
or  its  diameter. 

Hitherto  the  ratio  in  question  has  never  been  determined 
except  approximatively  ;  but  the  approximation  has  been  car- 
ried so  far,  that  a  knowledge  of  the  exact  ratio  would  afford 
no  real  advantage  whatever  beyond  that  of  the  approximate 
ratio.  Accordingly,  this  problem,  which  engaged  geometers 
so  deeply,  when  their  methods  of  approximation  were  less  per- 
fect, is  now  degraded  to  the  rank  of  those  idle  questions,  with 
which  no  one  possessing  the  slightest  tincture  of  geometrical 
science  will  oocupy  any  portion  of  his  time. 

Archimedes  showed  that  the  ratio  of  the  circumference  to 
the  diameter  is  included  between  3|f  and  3|f ;  hence  34  or 
2,f  affords  at  once  a  pretty  accurate  approximation  to  the  num- 
ber above  designated  by  n ;  and  the  simplicity  of  this  first  ap- 
proximation has  brought  it  into  very  general  use.  .  Metius, 
for  the  same  number,  found  the  much  more  accurate  value  fff. 
At  last  the  value  of  tt,  developed  to  a  certain  order  of  decimals, 
wasfound  by  other  calculators  to  be  3.1415926535897932,  &c. ; 


BOOK  V. 


133 


and  some  have  had  patience  enough  to  continue  these  decimals 
to  the  hundred  and  twenty-seventh,  or  even  to  the  hundred 
and  fortieth  place.  Such  an  approximation  is  evidently  equi- 
valent lo  perfect  correctness  :  the  root  of  an  imperfect  power 
is  in  no  case  more  accurately  known. 

The  following  problem  will  exhibit  one  of  the  simplest  ele- 
mentary methods  of  obtaining  those  approximations. 


i 


PROPOSITION  XIII.    PROBLEM. 

The  surface  of  a  regular  inscribed  polygon^  and  that  of  a  simi- 
lar polygon  circumscribed,  being  given ;  to  find  the  surfaces  of 
the  regular  inscribed  and  circumscribed  polygons  having 
double  the  number  of  sides. 

Let  AB  be  a  side  of  the  given 
inscribed  polygon ;  EF,  parallel  to 
AB,  a  side  of  the  circumscribed 
polygon  ;  C  the  centre  of  the  cir- 
cle. If  the  chord  AM  and  the 
tangents  AP,  BQ,  be  drawn,  AM 
will  be  a  side  of  the  inscribed 
polygon,  having  twice  the  num- 
ber of  sides;  andAP  +  PM=2PM 
or  PQ,  will  be  a  side  of  the  simi- 
lar circumscribed  polygon  (Prop. 
VI.  Cor.  3.).  Now,  as  the  same 
construction  will  take  place  at  each  of  the  angles  equal  to 
ACM,  it  will  be  sufficient  to  consider  ACM  by  itself,  the  tri- 
angles connected  with  it  being  evidently  to  each  other  as  the 
whole  polygons  of  which  they  form  part.  Let  A,  then,  be 
the  surface  of  the  inscribed  polygon  whose  side  is  AB,  B  that 
of  the  similar  circumscribed  polygon  ;  A'  the  surface  of  the 
polygon  whose  side  is  AM,  B'  .that  of  the  similar  circumscribed 
polygon :  A  and  B  are  given  ;  we  have  to  find  A^  and  B'. 

First.  The  triangles  ACD,  ACM,  having  the  common  ver- 
tex A,  are  to  each  other  as  their  bases  CD,  CM  ;  they  are  like- 
wise to  each  other  as  the  polygons  A  and  A,  of  which  they 
form  part :  hence  A  :  A  :  :  CD  :  CM.  Again,  the  triangles 
CAM,  CME,  having  the  common  vertex  M,  are  to  each  other 
as  their  bases  CA,  CE  ;  they  are  likewise  to  each  other  as  the 
polygons  A'  and  B  of  which  they  form  part ;  hence  A'  :  B  ;  : 
CA  :  CE.  But  since  AD  and  ME  are  parallel,  we  have 
CD  :  CM  :  ;  CA  :  CE;  hence  A  :  A'  :  :  A'  :  B  ;  hence  the 
polygon  A,  one  of  those  required,  is  a  mean  proportional  between 
the  two  given  polygons  A  and  F,and  consequently  A' =  V  A  x  B. 


124 


GEOMETRY. 


Secondly-  The  altitude  CM  be- 
ing common,  the  triangle  CPM  is 
to  the  triangle  CPE  as  PM  is  to 
PE  ;  but  since  CP  bisects  the  an- 
gle MCE,  we  have  PM  :  PE  :  : 
CM  :  CE  (Book  IV.  Prop. 
XVII.)  ;:CD  :  CA  :  :  A  :  A' : 
hence  CPM  :  CPE  :  :  A  :  A' ; 
and  consequently  CPM  :  CPM  + 
CPE  or  CME  :  :  A  :  A+ A'.  But 
CMPA,  or  2CMP,  and  CME  are 
to  each  other  as  the  polygons  B' 
and  B,  of  which  they  form  part :  hence  B'  :  B  :  :  2  A  :  A  +  A'. 
Now  A'  has  been  already  determined  ;  this  new  proportion  will 

serve  for  determining  B',  and  give  us  B'—       '     ;  and  thus  by 

A4"  A 
means  of  the  polygons  A  and  B  it  is  easy  to  find  the  polygons 
A'  and  B',  which  shall  have  double  the  number  of  sides. 


PROPOSITION  XIV.    PROBLEM. 

jTo  find  the  approximate  ratio  of  the  circumference  to  the 
diameter. 


Let  the  radius  Of  the  circle  be  1  ;  the  side  of  the  inscribed 
square  will  be  \/2  (Prop.  III.  Sch.),  that  of  the  circumscribed 
square  will  be  equal  to  the  diameter  2  ;  hence  the  surface  of 
the  inscribed  square  is  2,  and  that  of  the  circumsoribed  square 
is  4.  Let  us  therefore  put  Arr:2,  and  B=4  ;  by  the  last  pro- 
position we  shall  find  the  inscribed  octagon  A' =  \^  8 =2.8284271, 

1  R 

and  the  circumscribed  octagon  B'=„       /g=3.3137085.    The 

inscribed  and  the  circumscribed  octagons  being  thus  deter- 
mined, we  shall  easily,  by  means  of  them,  determine  the  poly- 
gons having  twice  the  number  of  sides.  We  have  only  in  this 
case  to  put  A=:2.8284271,  B  =  3.3137085;  we  shall  find  A'  = 

2A  B 

VA.B  =  3.0614674,and  B'=^^-^,  =  3.I825979.  These  poly- 
gons of  16  sides  will  in  their  turn  enable  us  to  find  the  polygons 
of  32  ;  and  the  process  may  be  continued,  till  there  remains 
no  longer  any  ditference  between  the  inscribed  and  the  cir- 
cumscribed polygon,  at  least  so  far  as  that  place  of  decimals 
where  the  computation  stops,  and  so  far  as  the  seventh  place, 
in  this  example.     Being  arrived  at  this  point,  we  shall  infer 


BOOK  V.  125 

that  the  last  result  expresses  the  area  of  the  circle,  which, 
since  it  must  always  lie  between  the  inscribed  and  the  circum- 
scribed polygon,  and  since  those  polygons  agree  as  far  as  a 
certain  place  of  decimals,  must  also  agree  with  both  as  far  as 
the  same  place. 

We  have  subjoined  the  computation  of  those  polygons,  car- 
ried on  till  they  agree  as  far  as  the  seventh  place  of  decimals. 

Number  of  sides  Inscribed  polygon.  Circumscribed  polygon. 

4 2.0000000  ....  4.0000000 

8 2.8284271  ....  3.3137085 

16  ....     .  3.0614674  ....  3.1825979 

32  ....     .  3.1214451  ....  3.1517249 

64 3.1365485  ....  3.1441184 

128 3.1403311  ....  3.1422236 

256 3.1412772  ....  «.1417504 

512 3.1415138  ....  3.1416321 

1024 3.1415729  ....  3.1416025 

2048  .....  3.1415877  ....  3.1415951 

4096 3.1415914  ....  3.1415933 

8192 3.1415923  ....  3.1415928 

16384  .....  3.1415925  ....  3.1415927 

32768 3.1415926  ....  3;1415926 

The  area  of  the  circle,  we  infer  therefore,  is  equal  to 
3.1415926.  Some  doubt  may  exist  perhaps  about  the  last  de- 
cimal figure,  owing  to  errors  proceeding  from  the  parts  omitted ; 
but  the  calculation  has  been  carried  on  with  an  additional 
figure,  that  the  final  result  here  given  might  be  absolutely  cor- 
rect even  to  the  last  decimal  place. 

Since  the  area  of  the  circle  is  equal  to  half  the  circumfe- 
rence multiplied  by  the  radius,  the  half  circumference  must  be 
3.1415926,  when  the  radius  is  1  ;  or  the  whole  circumference 
must  be  3.1415926,  when  the  diameter  is  1  :  hence  the  ratio 
of  the  circumference  to  the  diameter,  formerly  expressed  by  tt, 
is  equal  to  3.1415926. 


L* 


126  GEOMETRY. 


BOOK  VI. 


PLANES  AND  SOLID  ANGLES. 


Definitions, 

1.  A  straight  line  is  perpendicular  to  a  plane,  when  it  is  per- 
pendicular to  all  the  straight  lines  which  pass  through  its  foot 
in  the  plane.  Conversely,  the  plane  is  perpendicular  to  the 
line. 

The  foot  of  the  perpendicular  is  the  point  in  which  the  per- 
pendicular line  meets  the  plane. 

2.  A  line  is  parallel  to  a  plane,  when  it  cannot  meet  thai 
plane,  to  whatever  distance  both  be  produced.  Conversely, 
the  plane  is  parallel  to  the  line. 

3.  Two  planes  are  parallel  to  each  other,  when  they  cannot 
meet,  to  whatever  distance  both  be  produced. 

4.  The  angle  or  mutual  inclination  of  two  planes  is  the  quan- 
tity, greater  or  less,  by  which  they  separate  from  each  other ; 
this  angle  is  measured  by  the  angle  contained  between  two 
lines,  one  in  each  plane,  and  both  perpendicular  to  the  common 
intersection  at  the  same  point.  , 

This  angle  m.ay  be  acute,  obtuse,  or  a  right  angle. 
If  it  is  a  right  angle,  the  two  planes  are  perpendicular  to 
each  other. 

5.  A  solid  angle  is  the  angular  space  in-  g 
eluded  between  several  planes  which  meet  ^/f> 
at  the  same  point,                                                        yy    I  ^ 

Thus,  the  solid  angle  S,  is  formed  by  ^  y/  /  \ 
the  union  of  the  planes  ASB,  BSC,  CSD,         V^.......U^..)c 

DSA.  //"^  y 

Three  planes  at  least,  are  requisite  to      ^- v 

form  a  sohd  angle,  -A  B 


BOOK  VI.  127 

PROPOSITION  I.    THEOREM. 

A  straight  line  cannot  be  partly  in  aplane^  and  partly  out  of  it. 

For,  by  the  definition  of  a  plane,  when  a  straight  line  has 
two  points  common  with  a  plane,  it  lies  wholly  in  that  plane. 

Scholium.  To  discover  whether  a  surface  is  plane,  it  is  ne- 
cessary to  apply  a  straight  line  in  different  ways  to  that  sur- 
face, and  ascertain  if  it  touches  the  surface  throughout  its  whole 
extent. 

PROPOSITION  II.    THEOREM. 

Two  straight  lines,  which  intersect  each  other,  lie  in  the  same 
plane,  and  determine  its  position. 

Let  AB,  AC,  be  two  straight  lines' which 
intersect  each  other  in  A  ;  a  plane  may  be 
conceived  in  which  the  straight  line  AB  is 
found  ;  if  this  plane  be  turned  round  AB,  until 
it  pass  through  the  point  C,  then  the  line  AC, 
which  has  two  of  its  points  A  and  C,  in  this 
plane,  lies  wholly  in  it ;  hence  the  position  of 
the  plane  is  determined  by  the  single  condition  of  containing 
the  two  straight  lines  AB,  AC. 

Cor.  1.  A  triangle  ABC,  or  three  points  A,  B,  C,  not  in  a 
straight  line,  determine  the  position  of  a  plane. 

Cor.  2.  Hence  also  two  parallels 
AB,  CD,  determine  the  position  of  a 
plane ;  for,  drawing  the  secant  EF, 
the  plane  of  the  two  straight  lines 
AE,  EF,  is  that  of  the  parallels 
AB,  CD. 


I 


PROPOSITION  III.    THEOREM. 

ff  two  planes  cut  each  other,  their  common  intersection  will  be  a 
straight  line. 


128 


GEOMETRY. 


Let  the  two  planes  AB,  CD,  cut 
each  other.  Draw  the  straight  hne 
EF,  joining  any  two  points  E  and  F  in 
the  common  section  of  the  two  planes. 
This  line  will  lie  wholly  in  the  plane 
AB,  and  also  wholly  in  the  plane  CD 
(Book  I.  Def.  6.)  :  therefore  it  will  be 
in  both  planes  at  once,  and  conse- 
quently is  their  common  intersection. 


PROPOSITION  IV.    THEOREM. 

If  a  straight  line  he  perpendicular  io  two  straight  lines  at  their 
point  of  intersection,  it  will  be  perpendicular  to  the  plane  of 
those  lines. 


Let  MN  be  the  plane,  of  the 
two  hnes  BB,  CC,  and  let  AP 
be  perpendicular  to  them  at 
their  point  of  intersection  P ; 
then  will  AP  be  perpendicular 
to  every  line  of  the  plane  pass- 
ing through  P,  and  consequently 
to  the  plane  itself  (Def.  1.). 

Through  P,  draw  in  the  plane 
MN,  any  straight  line  as  PQ, 
and  through  any  point  of  this 
line,  as  Q,  draw  BQC,  so  that  BQ  shall  be  equal  to  QC  (Book 
IV.  Prob.  V.) ;  draw  AB,  AQ.  AC. 

The  base  BC  being  divided  into  two  equal  parts  at  the  point 
Q,  the  triangle  BPC  will  give  (Book  IV.  Prop.  XIV.), 
PC2+PB2=2PQ2+2QC2. 

The  triangle  BAC  will  in  like  manner  give, 
AC2+AB2=r2AQ2+2QC2. 

Taking  the  first  equation  from  the  second,  and  observing 
that  the  triangles  APC,  APB,  which  are  both  right  angled  at 

'  ^^^^       AC^— PC2= AP2,  and  AB^— PB2= AF ; 
we  shall  have 

AP2+AP2=2AQ2— 2PQ2. 
Therefore,  by  taking  the  halves  of  both,  we  have 
AP^=AQ2— PQ2,  or  AQ^^APHPQ^ ; 
hence  the  triangle  APQ  is  right  angled  at  P  ;  hence  AP  is  per- 
pendicular to  PQ. 


t 


BOOK  VI.  129 

Scholium.  Thus  it  is  evident,  not  only  that  a  straight  Hne 
may  be  perpendicular  to  all  the  straight  lines  which  pass 
through  its  foot  in  a  plane,  but  that  it  always  must  be  so,  when- 
ever it  is  perpendicular  to  two  straight  hues  drawn  in  the 
plane  ;  which  proves  the  first  Definition  to  be  accurate. 

Cor.  1.  The  perpendicular  AP  is  shorter  than  any  oblique 
line  AQ ;  therefore  it  measures  the  true  distance  from  the  point 
A  to  the  plane  MN. 

Cor.  2.  At  a  given  point  P  on  a  plane,  it  is  impossible  to 
erect  more  than  one  perpendicular  to  that  plane  ;  for  if  there 
could  be  two  perpendiculars  at  the  same  point  P,  draw  through 
these  two  perpendiculars  a  plane,  whose  intersection  with  the 
plane  MN  is  PQ ;  then  these  two  perpendiculars  would  be  per- 
pendicular to  the  line  PQ,  at  the  same  point,  and  in  the  same 
plane,  which  is  impossible  (Book  I.  Prop.  XIV.  Sch.). 

It  is  also  impossible  to  let  fall  from  a  given  point  out  of  a 
plane  two  perpendiculars  to  that  plane  ;  for  let  AP,  AQ,  be 
these  two  perpendiculars,  then  the  triangle  APQ  would  have 
two  right  angles  APQ,  AQP,  which  is  impossible. 


PHOPOSITION   V.    THEOREM. 

r     • 

If  from  a  point  without  a  plane,  a  perpendicular  he  drawn  to  the 
plane,  and  oblique  lines  he  drawn  to  different  points, 

\st.  Any  two  ohlique  lines  equally  distant  from  the  perpendicular 
will  he  equal. 

2d.  Of  any  two  ohlique  lines  unequally  distant  from  the  perpen- 
dicular, the  more  distant  will  he  the  longer. 

Let  AP  be  perpendicular  to 
the  plane  MN  ;  AB,  AC,  AD, 
oblique  lines  equally  distant 
from  the  perpendicular,  and 
AE  a  line  more  remote  :  then 
will  AB=AC=AD;  and  AE 
will  be  greater  than  AD. 

For,  the  angles  APB,  APC, 

APD,  being  right  angles,  if  we     

suppose  the  distances  PB,  PC,  IfJ 

PD,  to  be  equal  to  each  other,  the  triangles  APB,  APC,  APD, 
will  have  in  each  an  equal  angle  contained  by  two  equal  sides; 
therefore  they  will  be  equal ;  hence  the  hypothenuses,  or  the 
oblique  lines  AB,  AC,  AD,  will  be  equal  to  each  other.  In  like 

17 


130 


GEOMETRY. 


manner,  if  the  distance  PE  is  greater  than  PD  or  its  equal  PB, 
the  obHque  line  AE  will  evidently  be  greater  than  AB,  or  its 
equal  AD. 

Cor.  All  the  equal  oblique 
lines,  AB,  AC,  AD,  &c.  termi- 
nate in  the  circumference  BCD, 
described  from  Pthe  foot  of  the 
perpendicular  as  a  centre ; 
therefore  a  point  A  being  given 
out  of  a  plane,  the  point  P  at 
which  the  perpendicular  let  fall 
from  A  would  meet  that  plane, 
may  be  found  by  marking  upon 
that  plane  three  points  B,  C,  D,  equally  dist^t  from  the  pomt  A, 
and  then  finding  the  centre  of  the  circle  which  passes  through 
these  points  ;  this  centre  will  be  P,  the  point  sought. 

Scholium,  The  angle  ABP  is  called  the  inclination  of  the 
oblique  line  AB  to  the  plane  MN  ;  which  inclination  is  evidently 
equal  with  respect  to  all  such  lines  AB,  AC,  AD,  as  are  equally 
distant  from  the  perpendicular  ;  for  all  the  triangles  ABP,  ACP, 
ADP,  &c.  are  equal  to  each  other. 


PROPOSITION  VI.    THEOREM. 


If  from  a  point  without  a  plane,  a  perpendicular  he  let  fall  on  the 
plane,  and  from  the  foot  of  the  perpendicular  a  perpendicular 
he  drawn  to  any  line  of  the  plane,  and  from  the  point  of  inter- 
section a  line  he  drawn  to  the  first  point,  this  latter  line  will  he 
perpendicular  to  the  line  of  the  plane. 

Let  AP  be  perpendicular  to  the 
plane  NM,  and  PD  perpendicular  to 
BC ;  then  will  AD  be  also  perpen- 
dicular to  BC. 

Take  DB=DC,  and  draw  PB,  PC, 
AB,  AC.  Since  DB=DC,  the  ob- 
lique line  PB  — PC  :  and  with  regard 
to  the  perpendicular  AP,  since  PB  =  . 
PC,  the  oblique  line  AB  =  AC  (Prop. 
V.  Cor.)  ;  therefore  the  line  AD  has 
two  of  its  points  A  and  D  equally  distant  from  the  extremities 
B  and  C  ;  therefore  AD  is  a  perpendicular  to  BC,  at  its  middle 
point  D  (Book  I.  Prop.  XVI.  Cor.). 


EP' 


BOOK  VI.  181 

Cor.  It  is  evident  likewise,  that  BC  is  perpendicular  to  the 
plane  APD,  since  BC  is  at  once  perpendicular  to  the  two 
straight  lines  AD,  PD. 

Scholium.  The  two  lines  AE,  BC,  afford  an  instance  of  two 
lines  which  do  not  meet,  because  they  are  not  situated  in  the 
same  plane.  The  shortest  distance  between  these  lines  is  the 
straight  line  PD,  which  is  at  once  perpendicular  to  the  line  AP 
and  to  the  line  BC,  The  distance  PD  is  the  shortest  distance 
between  them,  because  if  we  join  any  other  two  points,  such 
as  A  and  B,  we  shall  have  AB>AD,  AD>PD;  therefore 
AB>PD. 

The  two  lines  AE,  CB,  though  not  situated  in  the  same  plane, 
are  conceived  as  forming  a  right  angle  with  each  other,  because 
AE  and  the  line  drawn  through  one  of  its  points  parallel  to 
BC  would  make  with  each  other  a  right  angle.  In  the  same 
manner,  the  line  AB  and  the  line  PD,  which  represent  any  two 
straight  Knes  not  situated  in  the  same  plane,  are  supposed  to 
form  with  each  other  the  same  angle,  which  would  be  formed 
by  AB  and  a  straight  line  parallel  to  PD  drawn  through  one 
of  the  points  of  AB. 


PROPOSITION  VII.    THEOREM. 

If  one  of  two  parallel  lines  he  perpendicular  to  a  plane,  the  other 
will  also  be  perpendicular  to  the  same  plane. 

Let  the  lines  ED,  AP,  be 
parallel ;  if  AP  is  perpen- 
dicular to  the  plane  NM, 
then  will  ED  be  also  per- 
pendicular to  it. 

Through  the  parallels  AP, 
DE,  pass  a  plane  ;  its  inter- 
section with  the  plane  MN 

will  be  PD^  in  the  plane  MN    "^ ^^ 

draw  BC  perpendicular  to  PD,  and  draw  AD. 

By  the  Corollary  of  the  preceding  Theorem,  BC  is  perpen- 
dicular to  the  pltuie  APDE  ;  therefore  the  angle  BDE  is  a  right 
angle ;  but  the  angle  EDP  is  also  a  right  angle,  since  AP  is 
perpendicular  to  PD,  and  DE  parallel  to  AP  (Book  I.  Prop. 
XX.  Cor.  1.)  ;  therefore  the  line  DE  is  perpendicular  to  the 
two  straight  lines  DP,  DB  ;  consequently  it  is  perpendicular  to 
their  plane  MN  (Prop.  IV.j. 


i 

K^ 

:p 

^ 

r 

l^S 


GEOMETRY. 


.. 

T 

7 

^ 

/c 

A 

N- 

Cor.  I.  Conversely,  if  the 
straight  lines  AP,  DE,  are 
perpendicular  to  the  same 
plane  MN,  they  will  be  par- 
allel ;  for  if  they  be  not  so, 
draw  through  the  point  D,  a 
line  parallel  to  AP,  this  par- 
allel will  be  perpendicular 
to  the  plane  MN  ;  therefore 
through  the  same  point  D  more  than  one  perpendicular  might 
be  erected  to  the  same  plane,  which  is  impossible  (Prop.  IV. 
Cor.  2.). 

Cor.  2.  Two  lines  A  and  B,  parallel  to  a  third  C,  are  par- 
allel to  each  other  ;  for,  conceive  a  plane  perpendicular  to  the 
line  C  ;  the  lines  A  and  B,  being  parallel  to  C,  will  be  perpen- 
dicular to  the  same  plane  ;  therefore,  by  the  preceding  Corol- 
lary, they  will  be  parallel  to  each  other. 

The  three  lines  are  supposed  not  to  be  in  the  same  plane  ; 
otherwise  the  proposition  would  be  already  known  (Book  I. 
Prop.  XXII.). 


PROPOSITION  VIII.    THEOREM. 

If  a  straight  line  is  parallel  to  a  straight  line  drawn  in  a  plane, 
it  will  hp  parallel  to  thai  plane. 


Let  AB  be  parallel  to  CD 
of  the  plane  NM  ;  then  will 
it  be  parallel  to  the  plane 
NM. 

For,  if  the  line  AB,  which 
lies  in  the  plane  ABDC, 
could  meet  the  plane  MN, 
this  could  only  be  in  some 


M 


:n 


point  of  the  line  CD,  the  common  intersection  of  the  two 
planes :  but  AB  cannot  meet  CD,  since  they  are  parallel ; 
hence  it  will  not  nieet  the  plane  MN ;  hence  it  is  parallel  to 
that  plane  (Def  2.). 


PROPOSITION  IX.    THEOREM. 


Two  planes  which  are  perpendicular  to  the  same  straight  line, 
are  parallel  to  each  other. 


BOOK  VI. 


133 


Let  the  planes  NM,  QP,  be  per-  - 
pendicular  to  the  Hne  AB,  then  will  P 
they  be  parallel. 

For,  if  they  can  meet  any  where, 
let  O  be  one  of  their  common 
points,  and  draw  OA,  OB  ;  the  line 
AB  which  is  perpendicular  to  the 
plane  MN,  is  perpendicular  to  the 
straight  line  OA  drawn  through  its  foot  in  that  plane  ;  for  the 
same  reason  AB  is  perpendicular  to  BO  ;  therefore  OA  and  OB 
are  two  perpendiculars  let  fall  from  the  same  point  O,  upon 
the  same  straight  line ;  which  is  impossible  (Book  I.  Prop.  XIV.); 
therefore  the  planes  MN,  PQ,  cannot  meet  each  other ;  conse- 
quently they  are  parallel. 


: ^ 

-A 

P            V 

M- 

\"J 

\ 

"\ 

PROPOSITION  X.    THEOREM. 

If  a  "plane  cut  two  parallel  planes ,  the  lines  of  intersection  will  he 

parallel. 


Let  the  parallel  planes  NM, 
QP,  be  intersected  by  the  plane 
EH  ;  then  will  the  lines  of  inter- 
section EF,  GH,  be  parallel. 

For,  if  the  Hues  EF,  GH,  lying 
in  the  same  plane,  were  not  par- 
allel, they  would  meet  each  other 
when  produced ;  therefore,  the 
planes  MN,  PQ,  in  which  those 
lines  lie,  would  also  meet ;  and 
hence  the  planes  would  not  be 
psirallel. 


M  E 


PROPOSITION  XI.    THEOREM. 


If  two  planes  are  parallel,  a  straight  line  which  is  perpendicular 
to  one,  is  also  perpendicular  to  the  other. 


M 


134 


iM 


OMETRY. 


Let  MN,  PQ,  be  two  parallel    •--. M 

planes,  and  let  AB  be  perpendicu-   P       '     ^ 


t 


X" 


^ 


I 


IT 


lar  to  NM  ;  then  will  it  also  be  per- 
pendicular to  QP. 

Having  drawn  any  line  BC  in 
the  plane  PQ,  through  the  lines  AB 
and  BC,  draw  a  plane  ABC,  inter- 
secting the  plane  MN  in  AD  ;  the 
intersection  AD  will  be  parallel  to  BC  (Prop.  X.) ;  but  the  line 
AB,  being  perpendicular  to  the  plane  MN,  is  perpendicular  to 
the  straight  line  AD ;  therefore  also,  to  its  parallel  BC  (Book 
I.  Prop.  XX.  Cor.  1.):  hence  the  line  AB  being  perpendicular 
to  any  line  BC,  drawn  through  its  foot  in  the  plane  PQ,  is  con- 
sequently perpendicular  to  that  plane  (Def.  1.). 


PROPOSITION  XII.    THEOREM. 

The  parallels  comprehended  between  two  parallel  planes  are 

equal. 


Let  MN,  PQ,  be  two  parallel 
planes,  and  FH,  GE,  two  paral- 
lel lines  :  then  will  EG=FH. 

For,  through  the  parallels  EG, 
FH,  draw  the  plane  EGHF,  in- 
tersecting the  parallel  planes  in 
EF  and  GH.    The  intersections 

EF,  GH,  are  parallel  to  each 
other  (Prop.  X.) ;  so  likewise  are 

EG,  FH  ;  therefore  the  figure 
EGHF  is  a  parallelogram ;  con- 
sequently, EG=FH. 


Cor.  Hence  it  follows,  that  two  parallel  planes  are  every 
where  equidistant :  for,  suppose  EG  were  perpendicular  to  the 
plane  PQ  ;  the  parallel  FH  would  also  be  perpendicular  to  it 
(Prop.  VH.),  and  the  two  parallels  would  likewise  be  perpen- 
dicular to  the  plane  MN  (Prop.  XI.)  ;  and  being  parallel,  they 
will  be  equal,  as  shown  by  the  Proposition. 


BOOK  VJ.i 


135 


PROPOSITION  XIII.    THEOREM. 


If  two  angles,  not  situated  in  the  same  plane^  have  their  sides 
parallel  and  lying  in  the  same  direction,  those  angles  will  he 
equal  and  their  planes  will  he  parallel. 

Let  the  angles  be  CAE  and  DBF. 

Make  AC=BD,  AEzr 
BF  ;  and  draw  CE,  DF, 
AB,  CD,  EF.  Since  AC 
is  equal  and  parallel  to 
BD,  the  figure  ABDC  is 
a  parallelogram  ;  therefore 
CD  is  equal  and  parallel 
to  AB.  For  a  similar  rea- 
son, EF  is  equal  and  par- 
allel to  AB ;  hence  also  CD 
is  equal  and  parallel  to 
EF  ;  hence  the  figure 
CEFD  is  a  parallelogram, 
and  the  side  CE  is  equal 
and  parallel  to  DF ;  therefore  the  triangles  CAE,  DBF,  have 
their  corresponding  sides  equal ;  therefore  the  angle  CAE= 
DBF. 

Again,  the  plane  ACE  is  parallel  to  the  plane  BDF.  For 
suppose  the  plane  drawn  through  the  point  A,  parallel  to  BDF, 
were  to  meet  the  lines  CD,  EF,  in  points  different  from  C  and 
E,  for  instance  in  G  and  H  ;  then,  the  three  lines  AB,  GD,  FH, 
would  be  equal  (Prop.  XII.) :  but  the  lines  AB,  CD,  EF,  are 
already  known  to  be  equal;  hence.  CD =GD,  and  FH=EF, 
which  is  absurd  ;  hence  the  plane  ACE  is  parallel  to  BDF. 

Cor,  If  two  parallel  planes  MN,  PQ  are  met  by  two  other 
planes  CABD,  EABF,  the  angles  CAE,  DBF,  formed  by  the 
mtersections  of  the  parallel  planes  will  be  equal ;  for,  the  inter- 
section AC  is  parallel  to  BD,  and  AE  to  BF  (Prop.  X.) ;  there- 
fore the  angle  CAE = DBF. 


PROPOSITION  XIV.    THEOREM. 


If  three  straight  lines,  not  situated  in  the  same  plane,  are  equal 
and  parallel,  the  opposite  triangles  formed  hy  joining  the  eX' 
tremities  of  these  lines  will  he  equal,  and  their  planes  will  he 
parallel. 


136 


GEOMETRY. 


Let  AB,  CD,  EF,  be  the 
lines. 

*  Since  AB  is  equal  and 
parallel  to  CD,  the  figure 
ABDC  is  a  parallelogram  ; 
hence  the  side  AC  is  equal 
and  parallel  to  BD.  For  a 
like  reason  the  sides  AE, 
BF,  are  equal  and  parallel, 
as  also  CE,  DF ;  therefore 
the  two  triangles  ACE,  BDF, 
are  equal ;  hence,  by  the  last 
Proposition,  their  planes  are 
parallel. 


PROPOSITION  XV.    THEOREM. 


If  two  straight  lines  he  cut  by  three  parallel  planes,  they  will  he 
divided  proportionally , 


Suppose  the  line  AB  to  meet 
the  parallel  planes  MN,  PQ, 
RS,  at  the  points  A,  E,  B ;  and 
the  line  CD  to  meet  the  same 
planes  at  the  points  C,  F,  D  : 
we  are  now  to  show  that 
AE  :  EB  :  :  CF  :  FD. 
Draw  AD  meeting  the  plane 
PQ  in  G,  and  draw  AC,  EG, 
GF,  BD ;  the  intersections  EG, 
BD,  of  the  parallel  planes  PQ, 
RS,  by  the  plane  ABD,  are 
parallel   (Prop.  X.)  ;  therefore 

AE  :  EB  :  :  AG  :  GD  ; 
in  like  manner,  the  intersections  AC,  GF,  being  parallel, 

AG  :  GD  :  :  CF  :  FD  ; 
the  ratio  AG  :  GD  is  the  same  in  both ;  hence 
AE  :  EB  :  :  CF  :  FD. 


H 

^ 

-  f 

\\ 

jsr 

+ 

M 

'  V 

.  /     W    " 

jl- 

— 

\ 

PROPOSITION  XVI.    THEOREM, 

If  a  line  is  perpendicular  to  a  plane,  every  plane  parsed  through 
.%  the  perpendicular,  will  also  he  perpendicular  to  the  plane. 


BOOK  VI. 


137 


Let  AP  be  perpendicular  to  the 
plane  NM  ;  then  will  every  plane 
passing  through  AP  be  perpendicu- 
lar to  NM. 

Let  BC  be  the  intersection  of  the 
planes  AB,  MN  ;  in  the  plane  MN, 
draw  DE  perpendicular  to  BP :  then 
the  line  AP,  being  perpendicular  to 
the  plane  MN,  will  be  perpendicu- 
lar to  each  of  the  two  straight  lines 
BC,  DE ;  but  the  angle  APD,  formed  by  the  two  perpendicu- 
lars PA,  PD,  to  the  common  intersection  BP,  measures  the 
angle  of  the  two  planes  AB,  MN  (Def.  4.) ;  therefore,  since  that 
angle  is  a  right  angle,  the  two  planes  are  perpendicular  to  each 
other. 

Scholium.  When  three  straight  lines,  such  as  AP,  BP,  DP, 
are  perpendicular  to  each  other,  each  of  those  lines  is  perpen- 
dicular to  the  plane  of  the  other  two,  and  the  three  planes  are 
perpendicular  to  each  other. 


PROPOSITION  XVII.    THEOREM. 

If  two  planes  are  perpendicular  to  each  other ^  a  line  draxon  in 
one  of  them  perpendicular  to  their  common  intersection^  will 
he  perpendicular  to  the  other  plane. 

Let  the  plane  AB  be  perpen- 
dicular, to  NM ;  then  if  the  line 
AP  be  perpendicular  to  the  inter- 
section BC,  it  will  also  be  perpen- 
dicular to  the  plane  NM. 

For,  in  the  plane  MN  draw  PD 
perpendicular  to  PB ;  then,  be- 
cause the  planes  are  perpendicu- 
lar, the  angle  APD  is  a  right  an- 
gle ;  therefore,  the  line  AP  is 
perpendicular  to  the  two  straight 
lines  PB,  PD  ;  therefore  it  is  perpendicular  to  their  plane  MN 
(Prop.  IV.). 

Cor.  If  the  plane  AB  is  perpendicular  to  the  plane  MN,  and 
if  at  a  point  P  of  the  common  intersection  we  erect  a  perpen- 
dicular to  the  plane  MN,  that  perpendicular  will  be  in  the  plane 
AB  ;  for,  if  not,  then,  in  the  plane  AB  we  might  draw  AP  per- 

M*  18 


138 


GEOMETRY. 


pendicular  to  PB  the  common  intersection,  and  this  AP,  at  the 
same  time,  would  be  perpendicular  to  the  plane  MN;  therefore 
at  the  same  point  P  there  would  be  two  perpendiculars  to  the 
plane  MN,  which  is  impossible  (Prop.  IV.  Cor.  2.). 


PROPOSITION  XVIII.    THEOREM. 


If  two  planes  are  perpendicular  to  a  third  plane,  their  common 
intersection  will  also  he  perpendicular  to  the  third  plane. 


'  Let  the  planes  AB,  AD,  be  per- 
pendicular to  NM ;  then  will  their 
intersection  AP  be  perpendicular 
toNM. 

For,  at  the  point  P,  erect  a  per- 
pendicular to  the  plane  MN  ;  that 
perpendicular  must  be  at  once  in 
the  plane  AB  and  in  the  plane  AD 
(Prop.  XVII.  Cor.)  ;  therefore  it 
is  their  common  intersection  AP. 


M 


M 


D 


^ 


y 


E 


iif 


PROPOSITION  XIX.    THEOREM. 


If  a  solid  angle  is  formed  by  three  plane  angles,  the  sum  of  any 
two  of  these  angles  will  he  greater  than  the  third. 

The  proposition  requires  demonstra- 
tion only  when  the  plane  angle,  which 
is  compared  to  the  sum  of  the  other 
two,  is  greater  than  either  of  them. 
Therefore  suppose  the  solid  angle  S  to 
be  formed  by  three  plane  angles  ASB, 
ASC,  BSC,  whereof  the  angle  ASB  is 
the  greatest;  we  are  to  show  that 
ASB<ASC  +  BSC. 

In  the  plane  ASB  make  the  angle  BSD=BSC,  draw  the 
straight  line  ADB  at  pleasure;  and  having  taken  SC  =  SD, 
draw  AC,  BC. 

The  two  sides  BS,  SD,  are  equal  to  the  two  BS,  SC  ;  the 
angle  BSD=:BSC  ;  therefore  the  triangles  BSD,  BSC,  are 
equal;  therefore  BD^BC.  ButAB<AC  +  BC;  taking  BD 
from  the  one  side,  and  from  the  other  its  equal  BC,  there  re- 


BOOK  VI.  139 


mains  AD<AC.  The  two  sides  AS,  SD,  are  equal  to  the 
two  AS,  SC  ;  the  third  side  AD  is  less  than  the  third  side  AC  ; 
therefore  the  angle  ASD<ASC  (Book  I.  Prop.  IX.  Sch.). 
Adding  BSD  =:BSC,  we  shall  have  ASD  +  BSD  or  ASB< 
ASC  +  BSC. 


PROPOSITION  XX.    THEOREM. 

TJie  sum  of  the  plane  angles  which  form  a  solid  angle  is  always 
less  than  four  right  angles. 

Cut  the  solid  angle  S  by  any  plane 
ABCDE ;  from  O,  a  point  in  that  plane, 
draw  to  the  several  angles  the  straight 
lines  AO,  OB,  OC,  OD,  OE. 

The  sum  of  the  angles  of  the  triangles 
ASB,  BSC,  &c.  formed  about  the  vertex 
S,  is  equal  to  the  sum  of  the  angles  of  an 
equal  number  of  triangles  AOB,  BOC,  &c. 
formed  about  the  point  O.  But  at  the 
point  B  the  sum  of  the  angles  ABO,  OBC, 
equal  to  ABC,  is  less  than  the  sum  of  the 
angles  ABS,  SBC  (Prop.  XIX.) ;  in  the  same  manner  at  the 
point  C  we  have  BCO  +  OCD<BCS  +  SCD;  and  so  with  all 
the  angles  of  the  polygon  ABCDE :  whence  it  follows,  that  the 
sum  of  all  the  angles  at  the  bases  of  the  triangles  whose  vertex 
is  in  O,  is  less  than  the  sum  of  the  angles  at  the  bases  of  the 
triangles  whose  vertex  is  in  S ;  hence  to  make  up  the  defi- 
ciency, the  sum  of  the  angles  formed  about  the  point  O,  is 
greater  than  the  sum  of  the  angles  formed  about  the  point  S. 
But  the  sum  of  the  angles  about  the  point  O  is  equal  to  four 
right  angles  (Book  I.  Prop.  IV.  Sch.) ;  therefore  the  sum  of  the 
plane  angles,  which  form  the  solid  angle  S,  is  less  than  four 
right  angles. 

Scholium.  This  demonstration  is  founded  on  the  supposition 
that  the  solid  angle  is  convex,  or  that  the  plane  of  no  one  sur- 
face produced  can  ever  meet  the  solid  angle  ;  if  it  were  other- 
wise, the  sum  of  the  plane  angles  would  no  longer  be  limited, 
and  might  be  of  any  magnitude. 

PROPOSITION  XXI.    THEOREM. 

If  two  solid  angles  are  contained  by  three  plane  angles  which  are 
equal  to  each  other ^  each  to  each,  the  planes  of  the  equal  angles 
will  he  equally  inclined  to  each  other* 


140  GEOMETRY. 

Let  the  angle  ASC==DTF,the 
angle  ASB-DTE,  and  the  an- 
gleBSC=ETF;  then  will  the 
inclination  of  the  planes  ASC, 
ASB,  be  equal  to  that  of  the 
planes  DTF,  DTE. 

Having  taken  SB  at  pleasure, 
draw  BO  perpendicular  to  the 
plane  ASC ;  from  the  point  O,  at  which  the  perpendicular 
meets  the  plane,  draw  OA,  OC  perpendicular  to  SA,  SC  ; 
draw  AB,  BC  ;  next  take  TE^SB  ;  draw  EP  perpendicular  to 
the  plane  DTF ;  from  the  point  P  draw  PD,  PF,  perpendicular 
respectively  to  TD,  TF ;  lastly,  draw  DE,  EF. 

The  triangle  SAB  is  right  angled  at  A,  and  the  triangle  TDE 
at  D  (Prop.  VI.) :  and  since  the  angle  ASB = DTE  we  have 
SBA=TED.  Likewise  SB=TE;  therefore  the  triangle  SAB 
is  equal  to  the  triangle  TDE;  therefore  SA=TD,  and  AB  =  DE. 
In  like  manner,  it  may  be  shown,  that  SC  =  TF,  and  BC  =  EF. 
That  granted,  the  quadrilateral  SAOC  is  equal  to  the  quadri- 
lateral TDPF:  for,  place  the  angle  ASC  upon  its  equal  DTF; 
because  SA=TD,  and  SC=TF,  the  point  A  will  fall  on  D, 
and  the  point  C  on  F ;  and  at  the  same  time,  AO,  which  is  per- 
pendicular to  SA,  will  fall  on  PD  which  is  perpendicular  to 
TD,  and  in  like  manner  OC  on  PF ;  wherefore  the  point  O 
will  fall  on  the  point  P,  and  AO  will  be  equal  to  DP.  But  the 
triangles  AOB,  DPE,  are  right  angled  at  O  and  P ;  the  hypo- 
thenuse  AB  =  DE,  and  the  side  AO=DP:  hence  those  trian- 
gles are  equal  (Book  I.  Prop.  XVII.)  ;  and  consequently,  the 
angle  OAB  =  PDE.  The  angle  OAB  is  the  inclination  of  the 
two  planes  ASB,  ASC  ;  and  the  angle  PDE  is  that  of  the  two 
planes  DTE,  DTF ;  hence  those  two  inclinations  are  equal  to 
each  other. 

It  must,  however,  be  observed,  that  the  angle  A  of  the  right 
angled  triangle  AOB  is  properly  the  inclination  of  the  two 
planes  ASB,  ASC,  only  when  the  perpendicular  BO  falls  on 
the  same  side  of  SA,  with  SC  ;  for  if  it  fell  on  the  other  side, 
the  angle  of  the  two  planes  would  be  obtuse,  and  the  obtuse 
angle  together  with  the  angle  A  of  the  triangle  OAB  would 
make  two  right  angles.  But  in  the  same  case,  the  angle  of  the 
two  planes  TDE,  TDF,  would  also  be  obtuse,  and  the  obtuse 
angle  together  with  the  angle  D  of  the  triangle  DPE,  would 
make  two  right  angles ;  and  the  angle  A  being  thus  always 
equal  to  the  angle  at  D,  it  would  follow  in  the  same  manner  that 
the  inclination  of  the  two  planes  ASB,  ASC,  must  be  equal  to 
that  of  the  two  planes  TDE,  TDF. 

Scholium.  If  two  solid  angles  are  contained  by  three  plane 


BOOK  VI.  141 

angles,  respectively  equal  to  each  other,  and  if  at  the  same  time 
the  equal  or  homologous  angles  are  disposed  in  the  same  man- 
ner in  the  two  solid  angles,  these  angles  will  be  equal  and  they 
will  coincide  when  applied  the  one  to  the  other.  We  have 
already  seen  that  the  quadrilateral  SAOC  may  be  placed  upon 
its  equal  TDPF ;  thus  placing  SA  upon  TD,  SC  falls  upon  TF, 
and  the  point  O  upon  the  point  P.  But  because  the  triangles 
AOB,  DPE,  are  equal,  OB,  perpendicular  to  the  plane  ASC, 
is  equal  to  PE,  perpendicular  to  the  plane  TDF  ;  besides,  those 
perdendiculars  lie  in  the  same  direction ;  therefore,  the  point 
B  will  fall  upon  the  point  E,  the  hne  SB  upon  TE,  and  the  two 
solid  angles  will  wholly  coincide. 

This  coincidence,  however,  takes  place  only  when  we  sup- 
pose that  the  equal  plane  angles  are  arranged  in  the  same  man- 
ner in  the  two  solid  angles  ;  for  if  they  were  arranged  in  an  in- 
verse order,  or,  what  is  the  same,  if  the  perpendiculars  OB,  PE, 
instead  of  lying  in  the  same  direction  with  regard  to  the  planes 
ASC,  DTF,  lay  in  opposite  directions,  then  it  would  be  impos- 
sible to  make  these  solid  angles  coincide  with  one  another.  It 
would  not,  however,  on  this  account,  be  less  true,  as  our  Theo- 
rem states,  that  the  planes  containing  the  equal  angles  must 
still  be  equally  inclined  to  each  other;  so  that  the  two  solid  an- 
gles would  be  equal  in  all  their  constituent  parts,  without, 
however,  admitting  of  superposition.  This  sort  of  equality, 
which  is  not  absolute,  or  such  as  admits  of  superposition,  de- 
serves to  be  distinguished  by  a  particular  liame  :  we  shall  call 
it  equality  by  symmetry. 

Thus  those  two  solid  angles,  which  are  formed  by  three 
plane  angles  respectively  equal  to  each  other,  but  disposed  in  an 
inverse  order,  will  be  called  angles  equal  by  symmetry ,  or  simply 
symmetrical  angles. 

The  same  remark  is  applicable  to  solid  angles,  which  are 
formed  by  more  than  three  plane  angles  :  thus  a  solid  angle, 
formed  by  the  plane  angles  A,  B,  C,  D,  E,  and  another  solid 
angle,  formed  by  the  same  angles  in  an  inverse  order  A,  E,  D, 
C,  B,  may  be  such  that  the  planes  which  contain  the  equal  an- 
gles are  equally  inclined  to  each  other.  Those  two  solid  angles, 
are  likewise  equal,  without  being  capable  of  superposition,  and 
are  called  solid  angles  equal  by  symmetry,  or  symmetrical  solid 
angles. 

Among  plane  figures,  equality  by  symmetry  does  not  pro- 
perly exist,  all  figures  which  might  take  this  name  being  abso- 
lutely equal,  or  equal  by  superposition  ;  the  reason  of  which  is, 
that  a  plane  figure  may  be  inverted,  and  the  upper  part  taken 
indiscriminately  for  the  under.  This  is  not  the  case  with  solids ; 
in  which  the  third  dimension  may  be  taken  in  two  different 
directions. 


142  GEOMETRY. 

BOOK  VII. 
POLYEDRONS. 


Definitions, 

1.  The  name  solid  polyedron,  or  simple  polyedron,  is  given 
to  every  solid  terminated  by  planes  or  plane  faces;  which 
planes,  it  is  evident,  will  themselves  be  terminated  by  straight 
lines. 

2.  The  common  intersection  of  two  adjacent  faces  of  a 
polyedron  is  called  the  side,  or  edge  of  the  polyedron. 

3.  The  prism  is  a  solid  bounded  by  several  parallelograms, 
which  are  terminated  at  both  ends  by  equal  and  parallel 
polygons. 

K  7c 


Be  if        O 

To  construct  this  solid,  let  ABODE  be  any  polygon  ;  then 
if  in  a  plane  parallel  to  ABODE,  the  lines  FG,  GH,  HI,  &c.  be 
drawn  equal  and  parallel  to  the  sides  AB,  BC,  CD,  &c.  thus 
forming  the  polygon  FGHIK  equal  to  ABODE  ;  if  in  the  next 
place,  the  vertices  of  the  angles  in  the  one  plane  be  joined  with 
the  homologous  vertices  in  the  other,  by  straight  lines,  AF,  BG, 
OH,  &c.  the  faces  ABGF,  BOHG,  <fec.  will  be  parallelograms, 
and  ABODE-K,  the  solid  so  formed,  will  be  a  prism. 

4.  The  equal  and  parallel  polygons  ABODE,  FGHIK,  are 
called  the  bases  of  the  prism;  the  parallelograms  taken  together 
constitute  the  lateral  or  convex  surface  of  the  prism ;  the  equal 
straight  lines  AF,  BG,  OH,  &c.  are  called  the  sides,  or  edges  of 
the  prism,  .J 

6.  The  altitude  of  a  prism  is  the  distance  between  its  two 
bases,  or  the  perpendicular  drawn  from  a  point  in  the  upper 
base  to  the  plane  of  the  lower  base.  .  ; ;    - 


BOOK  VII. 


143 


6.  A  prism  is  right,  when  the  sides  AF,  BG,  CH,  &c.  are 
perpendicular  to  the  planes  of  the  bases ;  and  then  each  of  them 
is  equal  to  the  ahitude  of  the  prism.  In  every  other  case  the 
prism  is  oblique,  and  the  altitude  less  than  the  side. 

7.  A  prism  is  triangular,  quadrangular,  pentagonal,  hex- 
agonal, &c.  when  the  base  is  a  triangle,  a  quadrilateral,  a 
pentagon,  a  hexagon,  &c. 

8.  A  prism  whose  base  is  a  parallelogram,  and 
which  has  all  its  faces  parallelograms,  is  named  a 
parallelopipedon. 

The  parallelopipedon  is  rectangular  when  all 
its  faces  are  rectangles. 

9.  Among  rectangular  parallelopipedons,  we 
distinguish  the  cube,  or  regular  hexaedron,  bounded 
by  six  equal  squares. 

10.  A  pyramid  is  a  solid  formed  by 
several  triangular  planes  proceeding  from 
the  same  point  S,  and  terminating  in  the 
different  sides  of  the  same  polygon 
ABCDE. 

The  polygon  ABCDE  is  called  the 
base  of  the  pyramid,  the  point  S  the 
vertex ;  and  the  triangles  ASB,  BSC, 
CSD,  &c.  form  its  convex  or  lateral  s\ir- 
face. 

11.  If  from  the  pyramid  S- ABCDE, 
the  pyramid  S-ahcde  be  cut  off  by  a 
plane  parallel  to  the  base,  the  remaining 
solid  ABCDE-rf,  is  called  a  truncated 
pyramid,  or  the  frustum  of  a  pyramid. 

12.  The  altitude  of  a  pyramid  is  the 
perpendicular  let  fall  from  the  vertex  upon 
base,  produced  if  necessary. 

13.  A  pyramid  is  triangular,  quadrangular,  (fee.  according 
as  its  base  is  a  triangle,  a  quadrilateral,  &c. 

14.  A  pyramid  is  regular,  when  its  base  is  a  regular  poly- 
gon, and  when,  at  the  same  time,  the  perpendicular  let  fall 
from  the  vertex  on  the  plane  of  the  base  passes  through  the 
centre  of  the  base.  That  perpendicular  is  then  called  the  axis 
of  the  pyramid. 

15.  Any  line,  as  SF,  drawn  from  the  vertex  S  of  a  regular 
pyramid,  perpendicular  to  either  side  of  the  polygon  which 
forms  its  base,  is  called  the  slant  height  of  the  pyramid. 

16.  The  diagonal  of  a  polyedron  is  a  straight  line  joining 
the  vertices  of  two  solid  angles  which  are  not  adjacent  to  each 
other. 


the  plane  of  the 


144 


GEOMETRY. 


17.  Two  polyedrons  are  similar  when  they  are  contained 
by  the  same  number  of  similar  planes,  similarly  situated,  and 
having  like  inclinations  with  each  other. 


PROPOSITION  I.     THEOREM. 


The  convex  surface  of  a  right  prism  is  equal  to  the  perimeter  of 
its  base  multiplied  by  its  altitude. 

Let  ABCDE-K  be  aright  prism :  then 
will  its  convex  surface  be  equal  to 
(AB  +  BC  +  CD  +  DE  +  EA)xAF. 

For,  the  convex  surface  is  equal  to  the 
sum  of  all  the  rectangles  AG,  BH,  CI, 
DK,  EF,  which  compose  it.  Now,  the 
altitudes  AF,  BG,  EH,  &c.  of  the  rect- 
angles, are  equal  to  the  altitude  of  the 
prism.  Hence,  the  sum  of  these  rectan- 
gles, or  the  convex  surface  of  the  prism, 
is  equal  to  (AB  +  BC  +  CD  +  DE  -f  E  A)  x 
AF ;  that  is,  to  the  perimeter  of  the  base  of  the  prism  multi- 
plied by  its  altitude. 

Cor,  If  two  right  prisms  have  the  same  altitude,  their  con- 
vex surfaces  will  be  to  each  other  as  the  perimeters  of  their 
bases. 


PROPOSITION  II.    THEOREM. 


In  every  prism,  the  sections  formed  by  parallel  planes,  are  equal 

polygons. 


Let  the  prism  AH  be  intersected  by 
the  parallel  planes  NP,  SV ;  then  are  the 
polygons  NOPQR,  STVXY  equal. 

For,  the  sides  ST,  NO,  are  parallel, 
being  the  intersections  of  two  parallel 
planes  with  a  third  plane  ABGF ;  these 
same  sides,  ST,  NO,  are  included  between 
the  parallels  NS,  OT,  which  are  sides  of 
the  prism:  hence  NO  is  equal  to  ST. 
For  like  reasons,  the  sides  OP,  PQ,  QR, 
&c.  of  the  section  NOPQR,  are  equal 
to  the  sides  TV,  VX,  XY,  ifec.  of  the  sec- 
tion STVXY,  each  to  each.     And  since 


BOOK  VII.  143 

the  equal  sides  are  at  the  same  time  parallel,  it  follows  that  the 
angles  NOP,  OPQ,  &c.  of  the  first  section,  are  equal  to  the 
angles  STV,TVX,  &c.  of  the  second,  each  to  each  (Book  VL 
Prop.  XIII.).  Hence  the  two  sections  NOPQR,  STVXY,  are 
equal  polygons. 

Cor.  Every  section  in  a  prism,  if  drawn  parallel  to  the  base, 
IS  also  equal  to  the  base. 


PROPOSITION  III.    THEOREM, 

If  a  pyramid  he  cut  hy  a  plane  parallel  to  its  base, 

IsL  The  edges  and  the  altitude  will  be  divided  proportionally, 

2d,  Tile  section  will  be  a  polygon  similar  to  the  base. 

Let  the  pyramid  S-ABCDE, 
of  which  SO  is  the  altitude, 
be  cut  by  the  plane  abcde  ; 
then  will  S«  :  SA  :  :  So  :  SO, 
and  the  same  for  the  other 
edges :  and  the  polygon  abcde, 
will  be  similar  to  the  base 
ABCDE. 

First.  Since  the  planes  ABC, 
abc,  are  parallel,  their  intersec- 
tions AB,  ab,  by  a  third  plane 
SAB  will  also  be  parallel 
(Book  VI.  Prop.  X.) ;  hence  the  triangles  SAB,  ^ab  are  simi* 
lar,  and  we  have  SA  :  Sa  :  :  SB  :  S6 ;  for  a  similar  reason, 
we  have  SB  :  Sft  :  :  SC  :  Sc;  and  so  on.  Hence  the  edges 
SA,  SB,  SC,  &c.  are  cut  proportionally  in  a,  b,  c,  &c.  The 
altitude  SO  is  likewise  cut  in  the  same  proportion,  at  the  point 
c ;  for  BO  and  bo  are  parallel,  therefore  we  have 
SO  :  So  :  :  SB  :  S&. 

Secondly.  Since  ab  is  parallel  to  AB,  be  to  BC,  cd  to  CD,  &c. 
the  angle  abc  is  equal  to  ABC,  the  angle  bed  to  BCD,  and  so  on 
(Book  VI.  Prop.  XIII.).  Also,  by  reason  of  the  similar  trian- 
gles SAB,  Sa6,  we  have  AB  :  «6  :  :  SB  :  S&  ;  and  by  reason 
of  the  similar  triangles  SBC,  S6c,  we  have  SB  :  S6  :  :  BC  : 
he  ;  hence  AB  :  ab  :  :  BC  :  be ;  we  might  likewise  have 
BC  :  he  :  :  CD  :  cd,  and  so  on.  Hence  the  polygons  ABCDE, 
abcde  have  their  angles  respectively  equal  and  their  homolo- 
gous sides  proportional ;  hence  they  are  similar. 

N  19 


146 


'     GEOMETRY. 


Cor.  1.  Let  S-ABCDE, 
S-XYZ  be  two  pyramids,  hav- 
ing a  common  vertex  and  the 
same  altitude.,  or  having  their 
bases  situated  in  the  same 
plane  ;  if  these  pyramids  are 
cut  by  a  plane  parallel  to  the 
plane  of  their  bases,  giving  the 
sections  abcde,  xyz,  then  will 
the  sections  ahcde,xyZfheto  each 
other  as  the  bases  ABCDE, 
XYZ. 

For,  the  polygons  ABCDE,  abcde,  being  similar,  their  sur- 
faces arenas  the  squares  of  the  homologous  sides  AB,  ab  ;  but 
AB  :  ab  :  :  SA  :  Sa;  hence  ABCDE  :  abcde  :  :  S A^  :  Sa^ 
For  the  same  reason,  XYZ  :  xyz  :  :  SX^  :  Sx^.  But  since 
abc  and  xyz  are  in  one  plane,  we  have  likewise  SA  :  Sot  :  : 
SX  :  Sx  (Book  VI.  Prop.  XV.) ;  hence  ABCDE  :  abcde  :  : 
XYZ  :  xyz  ;  hence  the  sections  abcdcy  xyz,  are  to  each  other 
as  the  bases  ABCDE,  XYZ. 

Cor,  2.  If  the  bases  ABCDE,  XYZ,  are  equivalent,  any  sec- 
tions abcde,  xyz,  made  at  equal  distances  from  the  bases,  will 
be  equivalent  likewise. 


PROPOSITION  IV.    THEOREM. 


The  convex  surface  of  a  regular  pyramid  is  equal  to  the  perime- 
ter of  its  base  multiplied  by  half  the  slant  height. 


For,  since  the  pyramid  is  regular,  the 
point  O,  in  which  the  axis  meets  the  base, 
is  the  centre  of  the  polygon  ABCDE 
(Def.  14.) ;  hence  the  lines  OA,  OB,  OC, 
&c.  drawn  to  the  vertices  of  the  base, 
are  equal. 

In  the  right  angled  triangles  S  AO,SBO, 
the  bases  and  perpendiculars  are  equal : 
hence  the  hypothenuses  are  equal :  and 
it  may  be  proved  in  the  same  way  that 
all  the  sides  of  the  right  pyramid  are 
equal.  The  triangles,  therefore,  which 
form  the  convex  surface  of  the  prism  are 
all  equal  to  each  other.  But  the  area  of 
either  of  these  triangles,  as  ESA,  is  equal 


BOOK  VII. 


147 


.  to  its  base  EA  multiplied  by  half  the  perpendicular  SF,  which 
is  the  slant  height  of  the  pyramid  :  hence  the  area  of  all  the  tri- 
angles, or  the  convex  surface  of  the  pyramid,  is  equal  to  the 
perimeter  of  the  base  multiplied  by  half  the  slant  height. 

Cor.  The  convex  surface  of  the  frustum  of  a  regular  pyra- 
mid is  equal  to  half  the  perimeters  of  its  upper  and  lower  bases 
multiplied  by  its  slant  height. 

For,  since  the  section  abcde  is  similar  to  the  base  (Prop.  III.), 
and  since  the  base  ABCDE  is  a  regular  polygon  (Def.  14.),  it 
follows  that  the  sides  ea,  ab,  be,  cd  and  de  are  all  equal  to  each 
other.  Hence  the  convex  surface  of  the  frustum  ABCDE-c? 
is  formed  by  the  equal  trapezoids  EA«e,  AB6«,  &c.  and  the 
perpendicular  distance  between  the  parallel  sides  of  either  of 
these  trapezoids  is  equal  to  F/j  the  slant  height  of  the  frustum. 
But  the  area  of  either  of  the  trapezoids,  as  AFiea,  is  equal  to 
^(EA-fea)  xF/  (Book  IV.  Prop.  VII.) :  hence  the  area  of  all 
ojf  them,  or  the  convex  surface  of  the  frustum,  is  equal  to  half 
the  perimeters  of  the  upper  and  lower  bases  multiplied  by  the 
slant  height,  .         . 

PROPOSITION  V.     THEOREM. 

-i 

If  the  three  planes  which  form  a  solid  angle  of  a  prism,  are  equal 
to  the  three  planes  which  form  the  solid  angle  of  another  prism, 
each  to  each,  and  are  like  siluatedy  the  two  prisms  will  be  equal 
to  each  other. 


Let  the  base  ABCDE  be  equal  to  the  base  abcde,  the  paral- 
lelogram ABGF  equal  to  the  parallelogram  abgf  and  the  par- 
allelogram BCHG  equal  to  bchg;  then  will  the  prism  ABCDE-K 
be  equal  to  the  prism  abcde-k. 

Tc 


For,  lay  the  base  ABCDE  upon  its  equal  abcde  ;  these  two 
—  will  coincide.     But  the  three  plane  angles  which  form 


^%. 


■^  . 


148 


GEOMETRY. 


the  solid  angle  B,  are  respectively  equal  to  the  three  plane 
angles,  which  form  the  solid  angle  h,  namely,  ABC  :=a&c, 
AEGrz=zahgy  and  GBC  —ghc  ;  they  are  also  similarly  situated  : 
hence  the  solid  angles  B  and  h  are  equal  (Book  VI.  Prop.  XXL 
Sch.) ;  and  therefore  the  side  BG  will  fall  on  its  equal  hg.  It 
is  likewise  evident,  that  by  reason  of  the  equal  parallelograms 
ABGF,  ahgf,  the  side  GF  will  fall  on  its  equal  gf,  and  in  the 
same  manner  GH  on  gh ;  hence,  the  plane  of  the  upper  base,. 
FGHIK  will  coincide  with  the  plane  fghik  (Book  VL  Prop.  II.). 


B        &  b        c 

But  the  two  upper  bases  being  equal  to  their  corresponding 
lower  bases,  are  equal  to  each  other :  hence  HI  will  coincide 
with  hi,  IK  with  ik,  and  KF  with  kf',  and  therefore  the  lateral 
faces  of  the  prisms  will  coincide :  therefore,  the  two  prisms 
coinciding  throughout  are  equal  (Ax.  13.). 

Cor.  Two  right  prisms,  which  have  equal  bases  and  equal  al- 
titudes, are  equal.  For,  since  the  side  AB  is  equal  to  ab,  and 
the  altitude  BG  to  bg,  the  rectangle  ABGF  will  be  equal  to 
abgf;  so  also  will  the  rectangle  BGHC  be  equal  to  bghc ;  and 
thus  the  three  planes,  which  form  the  solid  angle  B,  will  be 
equal  to  the  three  which  form  the  solid  angle  b.  Hence  the 
two  prisms  are  equal. 


PROPOSITION  VI.    THEOREM. 


In  every  parallelopipedon  the  opposite  planes  are  equal  and 

parallel. 


By  the  definition  of  this  solid,  the  bases 
ABCD,  EFGH,  are  equal  parallelogi-ams, 
and  their  sides  are  parallel :  it  remains 
only  to  show,  that  the  same  is  true  of  any 
two  opposite  lateral  faces^  such  as  AEHD, 
BFGC.  Now  AD  is  equal  and  parallel 
to  BC,  because  the  figure  ABCD  is  a  par- 


BOOK  VII.  149 

allelogram ;  for  a  like  reason,  AE  is  parallel  to  BF :  hence  the 
angle  DAE  is  equal  to  the  angle  CBF,  and  the  planes  DAE, 
CBF,  are  parallel  (Book  VI.  Prop.  XIII.)  ;  hence  also  the  par- 
allelogram DAEH  is  equal  to  the  parallelogram  CBFG.  In  the 
same  way.  it  might  be  shown  that  the  opposite  parallelograms 
ABFE,  DCGH,  are  equal  and  parallel. 

Cor.  1.  Since  the  parallelopipedon  is  asoHd  bounded  by  six 
planes,  whereof  those  lying  opposite  to  each  other  are  equal 
and  parallel,  it  follows  that  any  face  and  the  one  opposite  to  it, 
may  be  assumed  as  the  bases  of  the  parallelopipedon. 

Cor.  2.  The  diagonals  of  a  parallelopipedon  bisect  each 
other.  For,  suppose  two  diagonals  EC,  AG,  to  be  drawn  both 
through  opposite  vertices :  since  AE  is  equal  and  parallel  to 
CG,  the  figure  AEGC  is  a  parallelogram ;  hence  the  diagonals 
EC,  AG  will  nmtually  bisect  each  other.  In  the  same  manner, 
we  could  show  that  the  diagonal  EC  and  another  DF  bisect 
each  other ;  hence  the  four  diagonals  will  mutually  bisect  each 
other,  in  a  point  which  may  be  regarded  as  the  centre  of  the 
parallelopipedon. 

Scholium.  If  three  straight  lines  AB,  AE,  AD,  passing 
through  the  same  point  A,  and  making  given  angles  with  each 
other,  are  known,  a  parallelopipedon  may  be  formed  on  those 
lines.  For  this  purpose,  a  plane  must  be  passed  through  the 
extremity  of  each  line,  and  parallel  to  the  plane  of  the  other 
two  ;  that  is,  through  the  point  B  a  plane  parallel  to  DAE, 
through  D  a  plane  parallel  to  BAE,  and  through  E  a  plane 
parallel  to  BAD.  The  mutual  intersections  of  these  planes  will 
form  the  parallelopipedon  required. 


PROPOSITION  VII.    THEOREM. 

The  two  triangular  prisms  into  which  a  parallelopipedon  is  di- 
vided by  a  plane  passing  through  its  opposite  diagonal  edges, 
are  equivalent. 


N 


150  GEOMETRY. 

Let  the  parallelapipedon  ABCD-H  be  ^I 

divided  by  the  plane  BDHF  passing  through  ^^-^'^'TK 

its  diagonal  edges  :  then  will  the  triangular      "St^^^"^  /  \j\ 
prism  ABD-H  be  equivalent  to  the  trian-      ^Ci- — '7/''f\^G' 
gular  prism  BCD-H.  y\  L^CA^^ 

Through  the  vertices  B  and  F,  draw  the        /  "^^^'"^"^ 
planes  Ba6?c,  Yehg^  at  right  angles  to  the       /      L-^'-y^    / 
side  BF,the  former  meeting  AE,  DH,  CG,  aI^'^/     //A  / 
the  three  other  sides  of  the  parallelopipe-    ^vc'"[  / /\  Vrj 
don,  in  the  points  a,  rf,  c,  the  latter  in  e,  A,         xv/y^X^,^''^^ 
g :  the  sections  Ba^c,  Yehg,  will  be  equal         -r^^^^ —    ^ 
parallelograms.     They  are  equal,  because 
they  are  formed  by  planes  perpendicular  to  the  same  straight 
line,  and  consequently  parallel  (Prop.  II.) ;  they  are  parallelo- 
grams, because  «B,  dc,  two  opposite  sides  of  the  same  section, 
are  formed  by  the   meeting  of  one  plane  with  two  parallel 
planes  ABFE,  DCGH. 

For  a  like  reason,  the  figure  BaeF  is  a  parallelogram ;  so  also 
are  BF^c,  cdhg,  adhe,  the  other  lateral  faces  of  the  solid  ^adc-g', 
hence  that  solid  is  a  prism  (Def.  6.) ;  and  that  prism  is  right, 
because  the  side  BF  is  perpendicular  to  its  base. 

But  the  right  prism  ^adc-g,  is  divided  by  the  plane  BH  into 
the  two  right  prisms  B«cZ-A,  Bcc?-A,  since  they  have  the  same  alti- 
tude BF,  for  their  bases  B«(i,  Bc6?,  are  halves  of  the  same  par- 
allelogram (Prop.  V.  Cor.). 

It  is  now  to  be  proved  that  the  oblique  triangular  prism 
ABD-H  will  be  equivalent  to  the  right  triangular  prism  ^ad-h  ; 
and  since  those  prisms  have  a  common  part  ABD-/?,  it  will 
only  be  necessary  to  prove  that  the  remaining  parts,  namely, 
the  solids  BaADS,  FeEH^,  are  equivalent. 

Now,  by  reason  of  the  parallelograms  ABFE,  «BFe,  the 
sides  AE,  ae,  being  equal  to  their  parallel  BF,  are  equal  to  each 
other ;  and  taking  away  the  common  part  Ae,  there  remains 
AanEe.     In  the  same  manner  we  could  prove  J)d=Hh. 

Next,  to  bring  about  the  superposition  of  the  two  solids 
BaADc?,  FeEHA,  let  us  place  the  base  ¥eh  on  its  equal  BaeZ : 
the  point  e  falling  on  «,  and  the  point  h  on  d,  the  sides  eE,  AH, 
will  fall  on  their  equals  a  A,  cZD,  because  they  are  perpendicu- 
lar to  the  same  plane  Bad  Hence  the  two  sohds  in  question 
will  coincide  exactly  with  each  other  ;  hence  the  oblique  prism 
BAD-H,  is  equivalent  to  the  right  one  ^ad-h. 

In  the  same  manner  might  the  oblique  prism  BCD-H,  be 
proved  equivalent  to  the  right  prism  ^cd-h.  But  the  two  right 
prisms  Bad-h,  Bcd-?i,  are  equal,  since  they  have  the  same  alti- 
tude BF,  and  since  their  bases  Bad,  Bdcy  are  halves  of  the 
same  parallelogram  (Prop.  V.  Cor.).    Hence  the  two  trian- 


BOOK  VIL  15i 

gular  prisms  BAD-H,  BDC-G,  being  equivalent  to  the  equal 
right  prisms,  are  equivalent  to  each  other. 

Cor,  Every  triangular  prism  ABD-HEF  is  half  of  the  paral- 
lelopipedoii  AG  described  with  the  same  solid  angle  A,  and 
the  same  edges  AB,  AD,  AE. 


PROPOSITION  VIII.    THEOREM. 

If  two  parallelopipedons  have  a  common  base,  and  their  upper 
bases  in  the  same  plane  and  between  the  same  parallels,  thei/ 
will  be  equivalent. 

Let  the  parallelopipe- 
dons AG,  AL,  have  the 
common  base  AC,  and 
their  upper  bases  EG, 
MK,  in  the  same  plane, 
and  between  the  same 
parallels  HL,  EK ;  then 
will  they  be  equivalent. 

There  may  be  three 
cases,  according  as  EI  is        -^ 

greater,  less  than,  or  equal  to,  EF ;  but  the  demonstration  is 
the  same  for  all.  In  the  first  place,  then  we  shall  show  that 
the  triangular  prism  AEI-MDH,  is  equal  to  the  triangular 
prism  BFK-LCG. 

Since  AE  is  parallel  to  BF,  and  HE  to  GF,  the  angle  AEI=- 
BFK,  HEI = GFK,  and  HE  A = GFB.  Of  these  six  angles  the 
first  three  form  the  solid  angle  E,  the  last  three  the  solid  angle 
F ;  therefore,  the  plane  angles  being  respectively  equal,  and 
similarly  arranged,  the  solid  angles  F  and  E  must  be  equal 
(Book  VI.  Prop.  XXL).  Now,  if  the  prism  AEI-M  is  laid  on 
the  prism  BFK-L,  the  base  AEI  being  placed  on  the  base  BFK 
will  coincide  with  it  because  they  are  equal ;  and  since  the  solid 
angle  E  is  equal  to  the  solid  angle  F,  the  side  EH  will  fall  on 
its  equal  FG :  and  nothing  more  is  required  to  prove  the  coin- 
cidence of  the  two  prisms  throughout  their  whole  extent,  for 
the  base  AEI  and  the  edge  EH  determine  the  prism  AELM,  as 
the  base  BFK  and  the  edge  FG  determine  the  prism  BFK-L 
(Prop.  V.) ;  hence  these  prisms  are  equal. 

But  if  the  prism  AEI-M  is  taken  away  from  the  solid  AL, 
thei^e  will  remain  the  parallelopipedon  BADC-L ;  and  if  the 
prism  BFK-L  is  taken  away  from  the  same  solid,  there  will 
remain  the  parallelopipedon  BADC-G ;  hence  those  two  paral- 
lelopipedons BADC-L,  BADC-G,  are  equivalent. 


152 


GEOMETRY. 


PROPOSITION  IX.    THEOREM. 


I  >: 


Two  parallelopipedons,  having  the  same  base  and  the  same  alti' 
tude,  are  equivalent. 

Let  ABCD  be  the  com- 
mon base  of  the  two  par- 
allelopipedons  AG,  AL ; 
since  they  have  the  same 
altitude,  their  upper  bases 
EFGH,IKLM,willbein 
the  same  plane.  Also  the 
sides  EF  and  AB  will  be 
equal  and  parallel,  as  well 
as  IK  and  AB  ;  hence  EF 
is  equal  and  parallel  to 
IK;  for  a  like  reason,  GF 

is  equal  and  parallel  to  

LK.  Let  the  sides  EF,  GH,  be  produced,  and  likewise  KL, 
IM,  till  by  their  intersections  they  form  the  parallelogram 
NOPQ ;  this  parallelogram  will  evidently  be  equal  to  either 
of  the  bases  EFGH,  IKLM.  Now  if  a  third  parallelopipedon 
be  conceived,  having  for  its  lower  base  the  parallelogram 
ABCD,  and  NOPQ  for  its  upper,  the  third  parallelopipedon 
will  be  equivalent  to  the  parallelopipedon  AG,  since  with  the 
same  lower  base,  their  upper  bases  lie  in  the  same  plane 
and  between  the  same  parallels,  GQ,  FN  (Prop.  VIII.). 
For  the  same  reason,  this  third  parallelopipedon  will  also  be 
equivalent  to  the  parallelopipedon  AL ;  hence  the  two  paral- 
lelopipedons  AG,  AL,  which  have  the  same  base  and  the 
same  altitude,  are  equivalent. 


PROPOSITION  X.     THEOREM. 


Any  parallelopipedon  may  he  changed  into  an  equivalent  rectan- 
gular parallelopipedon  having  the  same  altitudt  and  an 
equivalent  base. 


BOOK  VII. 


153 


Let  AG  be  the  par- 
allelopipedon  proposed. 
From  the  points  A,  B,  C, 
D,drawAI,BK,CL,DM, 
perpendicular  tothe  plane 
of  the  base  ;  you  will  thus 
form  the  parallelopipe- 
don  AL  equivalent  to 
AG,  and  having  its  late- 
ral faces  AK,  BL,  &c, 
rectangles.  Hence  if  the 
base  ABCD  is  a  rectan- 
gle, AL  will  be  a  rectan- 
gular parallelopipedon  equivalent  to  AG,  and  consequently, 
the  parallelopipedon  required.  But  if  ABCD  is  not  a  rectangle, 
draw  AO  and  BN  perpendicular  to  CD,  and  mQ  IJ? 

OQ  and  NP  perpendicular  to  the  base  ;  you 
will  then  have  the  solid  ABNO-IKPQ,  which 
will  be  a  rectangular  parallelopipedon :  for 
by  construction,  the  bases  ABNO,  and  IKPQ 
are  rectangles  ;  so  also  are  the  lateral  faces, 
the  edges  AT,  OQ,  &c.  being  perpendicular 
to  the  plane  of  the  base  ;  hence  the  solid  AV 
is  a  rectangular  parallelopipedon.  But  the 
two  parallelopipedons  AP,  AL  may  be  con- 
ceived as  having  the  same  base  ABKI  and 
the  same  altitude  AO :  hence  the  parallelopipedon  AG,  which 
was  at  first  changed  into  an  equivalent  parallelopipedon  AL,^ 
is  again  changed  into  an  equivalent  rectangular  parallelopipe- 
don AP,  having  the  same  altitude  AI,  and  a  base  ABNO  equi- 
valent to  the  base  ABCD. 


PROPOSITION  XI.    THEOREM. 


Two  rectangular  parallelopipedons^  which  have  the  same  hase^ 
are  to  each  other  as  their  altitudes. 


154  GEOMETRY. 

Let  the  parallelopipedons  AG,  AL,  have  the  same  base  BD ; 
then  will  they  be  to  each  other  as  their  altitudes  AE,  AI. 

First,  suppose  the  altitudes  AE,  AI,  to  be    e  H 

to  each  other  as  two  whole  numbers,  as  15  is 
to  8,  for  example.     Divide  AE  into  15  equal        \F         I  \Cr 
parts ;  whereof  AI  will  contain  8  ;  and  through  q,  . 
Xf  y,  z,  &c.  the  points  of  division,  draw  planes     ■  -J 
parallel  to  the  base.     These  planes  will  cut      K  \       3M 
the  solid  AG  into  15  partial  parallelopipedons, 
all  equal  to  each  other,  because   they  have   ^X-    K 
equal  bases  and  equal  altitudes — equal  bases,  ^ '. 


smce  every  section  MIKL,  made  parallel  to  A\ 
the  base  ABCD  of  a  prism,  is  equal  to  that 
base  (Prop.  II.),  equal  altitudes,  because  the  -^ 
altitudes  are  the  equal  divisions  A:c,  xy^  yz, 
&c.  But  of  those  15  equal  parallelopipedons,  8  are  con- 
tained in  AL  ;  hence  the  solid  AG  is  to  the  soHd  AL  as  15  is  to 
8,  or  generally,  as  the  altitude  AE  is  to  the  ahitude  AI. 

Again,  if  the  ratio  of  AE  to  AI  cannot  be  exactly  expressed 
in  numbers,  it  is  to  be  shown,  that  notwithstanding,  we  shall 
have 

solid  AG  :  solid  AL  :  :  AE  :  AI. 
For,  if  this  proportion  is  not  correct,  suppose  we  have 

sol.  AG  :  sol.  AL  ;  :  AE  ;  AO  greater  than  AT. 
Divide  AE  into  equal  parts,  such  that  each  shall  be  less  than 
01 ;  there  will  be  at  least  one  point  of  divisiop  m,  between  O 
and  I.  Let  P  be  the  parallelopipedon,  whose  43ase  is  ABCD, 
and  altitude  Am  ;  since  the  altitudes  AE,  Awi,  are  to  each  other 
as  the  two  whole  numbers,  we  shall  have 

sol.  AG  :  P  :  :  AE  :  Am. 
But  by  hypothesis,  we  have 

«oZ.  AG  :  sol.  AL  :  :  AE  :  AO ; 
therefore,       f 

sol.AL  :  P  :  :  AO  :  Am. 
But  AO  is  greater  than  Am  ;  hence  if  the  proportion  is  correct, 
the  solid  AL  must  be  greater  than  P.     On  the  contrary,  how- 
ever, it  is  less :  hence  the  fourth  term  of  this  proportion 

soL  AG  ;  sol.  AL  :  :  AE  :  x, 
cannot  possibly  be  a  line  greater  than  AI.  By  the  same  mode 
of  reasoning,  it  might  be  shown  that  the  fourth  term  cannot  be 
less  than  AI ;  therefore  it  is  equal  to  AI ;  hence  rectangular 
parallelopipedons  having  the  same  base  are  to  each  other  as 
their  altitudes. 


BOOK  VII. 


155 


PROPOSITION  XII.    THEOREM. 

Two  rectangular  parallelopipedons,  having  the  same  altitude, 
are  to  each  other  as  their  bases. 


^ 


w 


\^ 


I 


V 


1 


Let  the  parallelopipedons      j-  '       E  IT 

AG,  AK,  have  the  same  al- 
titude AE ;  then  will  they  be 
to  each  other  as  their  bases 
AC,  AN. 

Having  placed  the  two  v     Ir hHCr 

solids  by  the  side  of  each  ^ 
other,  as  the  figure  repre- 
sents, produce  the  plane 
ONKL  till  it  meets  the 
plane  DCGH  in  PQ ;  you 
will  thus  have  a  third  par-  JM 
allelopipedon  AQ,  which 
may  be  compared  with  each  ^ 
of  the  parallelopipedons 
AG,  AK.  The  two  solids 
AG,  AQ,  having  the  same 
base  AEHD  are  to  each  other  as  their  altitudes  AB,  AO  ;  in 
like  manner,  the  two  solids  AQ,  AK,  having  the  same  base 
AOLE,  are  to  each  other  as  their  altitudes  AD,  AM.  Hence 
we  have  the  two  proportions, 

sol.  AG  :  sol.  AQ  :  :  AB  :  AO, 
sol.  AQ  :  sol.  AK  :  :  AD  :  AM. 
Multiplying  together  the  corresponding  terms  of  these  propor- 
tions, and  omitting  in  the  result  the  common  multiplier  sol.  AQ ; 
we  shall  have 

sol.AG  :  sol.AK  :  :  ABxAD  :  AOxAM. 
But  AB  X  AD  represents  the  base  ABCD  ;  and  AO  x  AM  rep- 
resents the  base  AMNO  ;  hence  two  rectangular  parallelopipe- 
dons of  the  same  altitude  are  to  each  other  as  their  bases. 


I 


PROPOSITION  XIII.    THEOREM. 


Any  two  rectangular  parallelopipedons  are  to  each  other  as  the 
products  of  their  bases  by  their  altitudes,  that  is  to  say,  as  the 
products  of  their  three  dimensions. 


166 


GEOMETRY. 


2S 


i 


K 


e 


For,  having  placed  the  two    I E H 

solids  AG,  AZ,  so  that  their 
surfaces  have  the  common 
angle  BAE,  produce  the 
planes   necessary   for   com-  Va 

pleting  the  third  parallelopi-  y  x 

pedon  AK  having  the  same 
altitude  with  the  parallelopi- 
pedon  AG.  By  the  last  propo- 
sition, we  shall  have 
sol.  AG  ;  sol.  AK  ;  : 
ABCD  :  AMNO.  kD 

But  the  two  parallelopipedons 
AK,  AZ,  having  the  same  base    -^ 
AMNO,  are  to  each  other  as  \ 

their  altitudes  AE,  AX ;  hence  :b 

we  have 

sol.  AK  :  sol.  AZ  :  :  AE  :  AX. 
Multiplying  together  the  corresponding  terms  of  these  propor- 
tions, and  omitting  in  the  result  the  common  multiplier  sol.  AK ; 
we  shall  have 

soZ.  AG  :  sol.  AZ  :  :  ABCDxAE  :  AMNO  x  AX. 

Instead  of  the  bases  ABCD  and  AMNO,  put  AB  x  AD  and 
AO  X  AM  it  will  give 

sol.AG  :  soLAZ  :  :  ABxADxAE  :  AOxAMxAX. 

Hence   any  two  rectangular    parallelopipedons    are   to   each 
other,  &c. 

Scholium.  We  are  consequently  authorized  to  assume,  as 
the  measure  of  a  rectangular  parallelopipedon,  the  product 
of  its  base  by  its  altitude,  in  other  words,  the  product  of  its 
three  dimensions. 

In  order  to  comprehend  the  nature  of  this  measurement,  it 
is  necessary  to  reflect,  that  the  number  of  linear  units  in  one 
dimension  of  the  base  multiplied  by  the  number  of  hnear  units 
in  the  other  dimension  of  the  base,  will  give  the  number  of 
superficial  units  in  the  base  of  the  parallelopipedon  (Book  IV. 
Prop.  IV.  Sch.).  For  each  unit  in  height  there  are  evidently 
as  many  sohd  units  as  there  are  superficial  units  in  the  base. 
Therefore,  the  number  of  superficial  units  in  the  base  multi- 
plied by  the  number  of  linear  units  in  the  altitude,  gives  the 
number  of  solid  units  in  the  parallelopipedon. 

If  the  three  dimensions  of  another  parallelopipedon  are 
valued  according  to  the  same  linear  unit,  and  multiplied  together 
in  the  same  manner,  the  two  products  will  be  to  each  ofier  as 


BOOK  VIL  157 

the  solids,  and  will   serve  to  express  their  relative  magni- 
tude. 

The  magnitude  of  a  solid,  its  volume  or  extent,  forms  what  is 
called  its  solidity ;  and  this  word  is  exclusively  employed  to 
designate  the  measure  of  a  solid  ;  thus  we  say  the  solidity  of  a 
rectangular  parallelopipedon  is  equal  to  the  product  of  its  base 
by  its  altitude,  or  to  the  product  of  its  three  dimensions. 

As  the  cube  has  all  its  three  dimensions  equal,  if  the  side  is 
1,  the  solidity  will  be  1  x  1  x  1  =:  1  :  if  the  side  is  2,  the  solidity 
will  be  2  X  2  X  2=8  ;  if  the  side  is  3,  the  solidity  will  be  3  x  3  x 
3=27 ;  and  so  on  :  hence,  if  the  sides  of  a  series  of  cubes  are 
to  each  other  as  the  numbers  1,  2,  3,  &c.  the  cubes  themselves 
or  their  solidities  will  be  as  the  numbers  1,  8,  27,  &c.  Hence 
it  is,  that  in  arithmetic,  the  cube  of  a  number  is  the  name  given 
to  a  product  which  results  from  three  factors,  each  equal  to 
this  number. 

If  it  were  proposed  to  find  a  cube  double  of  a  given  cube, 
the  side  of  the  required  cube  would  have  to  be  to  that  of  the 
given  one,  as  the  cube-root  of  2  is  to  unity.  Now,  by  a  geo- 
metrical construction,  it  is  easy  to  find  the  square  root  of  2  ; 
but  the  cube-root  of  it  cannot  be  so  found,  at  least  not  by  the 
simple  operations  of  elementary  geometry,  which  consist  in 
employing  nothing  but  straight  lines,  two  points  of  which  are 
known,  and  circles  whose  centres  and  radii  are  determined. 

Owing  to  this  difficulty  the  problem  of  the  duplication  of 
the  cube  became  celebrated  among  the  ancient  geometers,  as 
well  as  that  of  the  trisection  of  an  angle,  which  is  nearly  of  the 
same  species.  The  solutions  of  which  such  problems  are  sus- 
ceptible, have  however  long  since  been  discovered  ;  and  though 
less  simple  than  the  constructions  of  elementary  geometry,  they 
are  not,  on  that  account,  less  rigorous  or  less  satisfactory. 


PROPOSITION  XIV.    THEOREM. 

The  solidity  of  a  parallelopipedon,  and  generally  of  any  prism, 
is  equal  to  the  product  of  its  base  by  its  altitude. 

For,  in  the  first  place,  any  parallelopipedon  is  equivalent  to 
a  rectangular  parallelopipedon,  having  the  same  altitude  and 
an  equivalent  base  (Prop.  X.).  Now  the  solidity  of  the  latter 
is  equal  to  its  base  multiplied  by  its  height ;  hence  the  solidity 
of  the  former  is,  in  like  manner,  equal  to  the  product  of  its  base 
by  its  altitude. 

In  the  second  place,  any  triangular  prism  is  half  of  the  par- 
allelopipedon so  constructed  as  to  have  the  same  altitude  and 
a  double  base  (Prop.  VII.).  But  the  solidity  of  the  latter  is  equal 


158 


GEOMETRY. 


to  its  base  multiplied  by  its  altitude ;  hence  that  of  a  triangular 
prism  is  also  equal  to  the  product  of  its  base,  which  is  half  that 
of  the  parallelopipedon,  multiplied  into  its  altitude. 

In  the  third  place,  any  prism  may  be  divided  into  as  many 
triangular  prisms  of  the  same  altitude,  as  there  are  triangles 
capable  of  being  formed  in  the  polygon  which  constitutes  its 
base.  But  the  solidity  of  each  triangular  prism  is  equal  to  its 
base  multiplied  by  its  altitude  ;  and  since  the  altitude  is  the 
same  for  all,  it  follows  that  the  sum  of  all  the  partial  prisms 
must  be  equal  to  the  sum  of  all  the  partial  triangles,  which  con- 
stitute their  bases,  multiplied  by  the  common  altitude. 

Hence  the  solidity  of  any  polygonal  prism,  is  equal  to  the 
product  of  its  base  by  its  altitude. 

Co7\  Comparing  two  prisms,  which  have  the  same  altitude, 
the  products  of  their  bases  by  their  altitudes  will  be  as  the 
bases  simply  ;  hence  two  prisms  of  the  same  altitude  are  to  each 
other  as  their  bases.  For  a  like  reason,  two  prisms  of  the  same 
base  are  to  each  other  as  their  altitudes.  And  when  neither  their 
bases  nor  their  altitudes  are  equal,  their  solidities  will  be  to 
each  other  as  the  products  of  their  bases  and  altitudes. 

PROPOSITION  XV.     THEOREM. 

Tivo  tjnangucar  pyramids,  having  equivalent  bases  and  equal 
altitudes f  are  equivalent,  or  equal  in  solidity. 


liCt  S-ABC,  S-«6c,  be  those  two  pyramids  ;  let  their  equiva- 
lent bases  ABC,  abc,  be  situated  in  the  same  plane,  and  let  AT 
be  their  common  altitude.  If  they  are  not  equivalent,  let  ^-abc 


BOOK  VII.  159 

be  the  smaller :  and  suppose  Ka  to  be  the  altitude  of  a  prism, 
which  having  ABC  for  its  base,  is  equal  to  their  difference. 

Divide  the  altitude  AT  into  equal  parts  Ax,  xy,  yz,  &c.  each 
less  than  Aa,  and  let  k  be  one  of  those  parts  ;  through  the  points 
of  division  pass  planes  parallel  to  the  plane  of  the  bases  ;  the 
corresponding  sections  formed  by  these  planes  in  the  tw^o  pyra- 
mids w^ill  be  respectively  equivalent,  namely  DEF  to  def,  GHI 
io  ghi,  &c,  (Prop.  III.  Cor.  2.). 

This  being  granted,  upon  the  triangles  ABC,  DEF,  GHI,  &c. 
taken  as  bases,  construct  exterior  prisms  having  for  edges  the 
parts  AD,  DG,  GK,  &c.  of  the  edge  SA  ;  in  like  manner,  on 
bases  def,  ghi,  klm,  &c.  in  the  second  pyramid,  construct  inte- 
rior prisms,  having  for  edges  the  corresponding  parts  of  Sa. 
It  is  plain  that  the  sum  of  all  the  exterior  prisms  of  the  pyramid 
S-ABC  will  be  greater  than  this  pyramid  ;  and  also  that  the 
sum  of  all  the  interior  prisms  of  the  pyramid  S-abc  will  be  less 
than  this  pyramid.  Hence  the  difference,  between  the  sum  of  all 
the  exterior  prisms  and  the  sum  of  all  the  interior  ones,  must  be 
greater  than  the  difference  between  the  two  pyramids  tliem- 
selves. 

Now,  beginning  with  the  bases  ABC,  ahcj  the  second  exte- 
rior prism  DEF-G  is  equivalent  to  the  first  interior  prism  def-ay 
because  they  have  the  same  altitude  k,  and  their  bases  DEF, 
def,  are  equivalent ;  for  like  reasons,  the  third  exterior  prism 
GHI-K  and  the  second  interior  prism  ghi-d  are  equivalent ; 
the  fourth  exterior  and  the  third  interior  ;  and  so  on,  to  the  last 
in  each  series.  Hence  all  the  exterior  prisms  of  the  pyramid 
S-ABC,  excepting  the  first  prism  ABC-D,  have  equivalent  cor- 
responding ones  in  the  interior  prisms  of  the  pyramid  S-abc : 
hence  the  prism  ABC-D,  is  the  difference  between  the  sum  of 
all  the  exterior  prisms  of  the  pyramid  S-ABC,  and  the  sum  of 
the  interior  prisms  of  the  pyramid  S-abc.  But  the  difference 
between  these  two  sets  of  prisms  has  already  been  proved  to 
be  greater  than  that  of  the  two  pyramids  ;  which  latter  diffe- 
rence we  supposed  to  be  equal  to  the  prism  «-ABC  :  hence  the 
prism  ABC-D,  must  be  greater  than  the  prism  a- ABC.  But  in 
reality  it  is  less  ;  for  they  have  the  same  base  ABC,  and  the 
altitude  Ax  of  the  first  is  less  than  Aa  the  altitude  of  the  second. 
Hence  the  supposed  inequality  between  the  two  pyramids  can- 
not exist ;  hence  the  two  pyramids  S-ABC,  S-abc,  having  equal 
altitudes  and  equivalent  bases,  are  themselves  equivalent. 


16a 


GEOMETRY. 


PROPOSITION  XVI.     THEOREM. 

Every  triangular  pyramidis  a  third  part  of  the  triangular  pt'ism 
having  the  same  base  and  the  same  altitude. 


Let  F-ABC  be  a  triangular 
pyramid,  ABC-DEF  a  triangular 
prism  of  the  same  base  and  the 
same  altitude  ;  the  pyramid  will 
be  equal  to  a  third  of  the  prism. 

Cut  off  the  pyramid  F-ABC 
from  the  prism,  by  the  plane 
FAC  ;  there  will  remain  the  solid 
F-ACDE,  which  may  be  consi- 
dered as  a  quadrangular  pyramid, 
whose  vertex  is  F,  and  whose  base 
is  the  parallelogram  ACDE. 
Draw  the  diagonal  CE  ;  and  pass 
the  plane  FCE,  which  will  cut  the 
quadrangular  pyramid  into  two  triangular  ones  F-ACE,F-CDE. 
These  two  triangular  pyramids  have  for  their  common  altitude 
the  perpendicular  let  fall  from  F  on  the  plane  ACDE ;  they 
have  equal  bases,  the  triangles  ACE,  CDE  being  halves  of  the 
same  parallelogram  ;  hence  the  two  pyramids  F-ACE,  F-CDE, 
are  equivalent  (Prop.  XV.).  But  the  pyramid  F-CDE  and  the 
pyramid  F-ABC  have  equal  bases  ABC,DEF ;  they  have  also  the 
same  altitude,  namely,  the  distance  between  the  parallel  planes 
ABC,  DEF ;  hence  the  two  pyramids  are  equivalent.  Now  the 
pyramid  F-CDE  has  already  been  proved  equivalent  to  F-ACE  ; 
hence  the  three  pyramids  F-ABC,  F-CDE,  F-ACE,  which 
compose  the  prism  ABC-DEF  are  all  equivalent.  Hence  the 
pyramid  F-ABC  is  the  third  part  of  the  prism  ABC-DEF,  which 
has  the  same  base  and  the  same  altitude. 

Cor.  The  solidity  of  a  triangular  pyramid  is  equal  to  a  third 
part  of  the  product  of  its  base  by  its  altitude. 


PROPOSITION  XVII.    THEOREM. 


The  solidity  of  every  pyramid  is  equal  to  the  base  multiplied  by 
a  third  of  the  altitude. 


BOOK  VII.  161 

Let  S-ABCDE  be  a  pyramid. 

Pass  the  planes  SEE,  SEC,  through  the 
diagonals  EB,  EC ;  the  polygonal  pyramid 
S-ABCDE  will  be  divided  into  several  trian- 

fular  pyramids  all  having  the  same  altitude 
10.  But  each  of  these  pyramids  is  measured 
by  multiplying  its  base  ABE,  BCE,  or  CDE, 
by  the  third  part  of  its  altitude  SO  (Prop.  XVI. 
Cor.) ;  hence  the  sum  of  these  triangular  pyra- 
mids, or  the  polygonal  pyramid  S-ABCDE 
will  be  measured  by  the  sum  of  the  triangles 
ABE,  BCE,  CDE,  or  the  polygon  ABODE, 
multiplied  by  one  third  of  SO  ;  hence  every  pyramid  is  mea- 
sured by  a  third  part  of  the  product  of  its  base  by  its  altitude. 

Cor.  1.  Every  pyramid  is  the  third  part  of  the  prism  which 
has  the  same  base  and  the  same  altitude. 

Cor.  2.  Two  pyramids  having  the  same  altitude  are  to  each 
other  as  their  bases. 

Cor.  3.  Two  pyramids  havmg  equivalent  bases  are  to  each 
other  as  their  altitudes. 

Cor.  4.  Pyramids  are  to  each  other  as  the  products  of  their 
bases  by  their  altitudes. 

Scholium.  The  solidity  of  any  polyedral  body  may  be  com- 
puted, by  dividing  the  body  into  pyramids ;  and  this  division 
may  be  accomplished  in  various  ways.  One  of  the  simplest 
is  to  make  all  the  planes  of  division  pass  through  the  vertex 
of  one  solid  angle  ;  in  that  case,  there  will  be  formed  as  many 
partial  pyramids  as  the  polyedron  has  faces,  minus  those  faces 
which  form  the  solid  angle  whence  the  planes  of  division 
proceed. 


PROPOSITION  XVIII.    THEOREM. 

If  a  pyramid  he  cut  by  a  plane  parallel  to  its  base,  the  fmsium 
that  remains  wlien  the  small  pyramid  is  taken  away,  iV  equi- 
valent to  the  sum  of  three  pyramids  having  for  their  common 
altitude  the  altitude  of  the  fi^stum,  and  for  bases  the  lower 
base  of  the  frustum,  the  upper  base,  and  a  mean  proportional 
between  the  two  bases. 

O*  21 


162 


GEOMETRY. 


Let  S-ABCDE  be  a  pyra- 
mid cut  by  the  plane  abcdet 
parallel  to  its  base;  let  T-FGH 
be  a  triangular  pyramid  hav- 
ing the  same  altitude  and  an 
equivalent  base  with  the  pyra- 
mid S-ABCDE.  The  two 
bases  may  be  regarded  as 
situated  in  the  same  plane  ;  in 
which  case,  the  plane  ahcdy  if 
produced,  will  form  in  the  triangular  pyramid  a  section  fgh 
situated  at  the  same  distance  above  the  common  plane  of  the 
bases  ;  and  therefore  the  section  ^A  will  be  to  the  section  abode 
as  the  base  FGH  is  to  the  base  ABD  (Prop.  III.),  and  since 
the  bases  are  equivalent,  the  sections  will  be  so  hkewise. 
Hence  the  pyramids  S-abcde,  T-fgh  are  equivalent,  for  their 
altitude  is  the  same  and  their  bases  are  equivalent.  The  whole 
pyramids  S-ABCDE,  T-FGH  are  equivalent  for  the  same  rea- 
son ;  hence  the  frustums  ABD-dahf  FGH-^  are  equivalent ; 
hence  if  the  proposition  can  be  proved  in  the  single  case  of 
the  frustum  of  a  triangular  pyramid,  it  will  be  true  of  every 
other.  • 

I^et  FGH-/i^  be  the  frustum  of  a  tri- 
angular pyramid,  having  parallel  bases : 
through  the  three  points  F,  g,  H,  pass 
the  plane  F^H ;  it  will  cut  off  from  the 
frustum  the  triangular  pyramid  ^-FGH. 
This  pyramid  has  for  its  base  the  lower 
base  FGH  of  the  frustum  ;  its  altitude 
likewise  is  that  of  the  frustum,  because 
the  vertex  g  lies  in  the  plane  of  the  up- 
per base  fgh. 

This  pyramid  being  cut  off,  there  will 
remain  the  quadrangular  pyramid 
g-fhUF,  whose  vertex  is  g,  and  base  fhHF,  Pass  the  plane 
j^H  through  the  three  points  /,  ^,  H ;  it  will  divide  the  quad- 
rangular pyramid  into  two  triangular  pyramids  g-FfH,  g-fW. 
The  latter  has  for  its  base  the  upper  base  gfli  of  the  frustum ; 
and  for  its  altitude,  the  altitude  of  the  frustum,  because  its  ver- 
tex H  lies  in  the  lower  base.  Thus  we  already  know  two  of 
the  three  pyramids  which  compose  the  frustum. 

It  remains  to  examine  the  third  g-FfH.  Now,  if  ^K  be 
drawn  parallel  to  /F,  and  if  we  conceive  a  new  pyramid 
K-FfH,  having  K  for  its  vertex  and  FfH  for  its  base,  these 
two  pyramids  will  have  the  same  base  FfH ;  they  will  also 
have  the  sarte  altitude,  because  their  vertices  g  and  K  lie  in 
the  line  ^K,  parallel  to  F/,  and  consequently  parallel  to  the 


BOOK  TIL 


163 


plane  of  the  base  :  hence  these  pyramids  are  equivalent.  But 
the  pyramid  K-F/H  may  be  regarded  as  having  its  vertex  iu 
/,  and  thus  its  altitude  will  be  the  same  as  that  of  the  frustum : 
as  to  its  base  FKH,  we  are  now  to  show  that  this  is  a  mean 
proportional  between  the  bases  FGH  and  fgh.  Now,  the  tri- 
angles FHK,^A,  have  each  an  equal  angle  F=/;  hence 

FHK  :  /^A  :  :  FKxFH  :  fgxfh  (Book  IV.  Prop.  XXIV.)  ; 
but  because  of  the  parallels,  FK=^,  hence 
FHK  ifgh  :  :  FH  :  fh. 
We  have  also, 

FHG  :  FHK  :  :  FG  :  FK  or  fg. 
But  the  similar  triangles  FGH,  fgh  give 

FG-.fg-.-.FH-.fli; 
hence, 

FGH  :  FHK  :  :  FHK  :  fgh; 
or  the  base  FHK  is  a  mean  proportional  between  the  two 
bases  FGH,  fgh.  Hence  the  frustum  of  a  triangular  pyramid 
is  equivalent  to  three  pyramids  whose  common  altitude  is  that 
of  the  frustum  and  whose  bases  are  the  lower  base  of  the 
frustum,  the  upper  base,  and  a  mean  proportional  between  the 
two  bases. 


PROPOSITION  XIX.     THEOREM. 

Similar  triangular  prisms  are  to  each  other  as  the  cubes  ofthett 
homologous  sides. 


Let  CBD-P,  chd'p,  be  two 
similar  triangular  prisms,  of 
which  BC,  he,  are  homologous 
sides:  then  will  the  prism 
CBD-P  be  to  the  prism  chd-p, 
asBC3to6c3. 

For,  since  the  prisms  are 
similar,  the  planes  which  con- 
tain the  homologous  solid  an-  C  C  B 
gles  B  and  h,  are  similar,  like  placed,  and  equally  inclined  to 
each  other  (Def.  1 7.) :  hence  the  solid  angles  B  and  h,  are  equal 
(Book  VI.  Prop.  XXL  Sch.).  If  these  solid  angles  be  applied 
to  each  other,  the  angle  chd  will  coincide  with  CBD,  the  side  ha 
with  B  A,  and  the  prism  chd-p  will  take  the  position  ^cd-p.  From  A 
draw  AH  perpendicular  to  the  common  base  of  the  prisms :  then 
will  the  plane  BAH  be  perpendicular  to  the  plane  of  the  com- 


164 


GEOMETRY. 


mon  base  (Book  VI.  Prop.  XVI.).  Through  a,  in  the  plane  BAH, 
draw  ah  perpendicular  to 
BH  :  then  will  ah  also  be  per- 
pendicular to  the  base  BDC 
(Book  VI.  Prop.  XVII.)  ;  and 
AH,  ah  will  be  the  altitudes 
of  the  two  prisms. 

Now,  because  of  the  similar 
triangles  ABH,aBA,  and  of  the 
similar  parallelograms  AC,  «c, 
we  have 

AH  :  a/t  :  :  AB  :  «&  :  :  BC  :  he. 
But  since  the  bases  are  similar,  we  have 
base  BCD  :  base  bed  :  :  BC^  :  be^  (Book  IV.  Prop.  XXV.)  ; 
hence, 

base  BCD  :  base  bed  :  :  AH^  :  ah\ 
Multiplying  the  antecedents  by  AH,  and  the  consequents  by 
ahy  and  we  have 

base  BCD  x  AH  :  base  bed  xah  :  :  AiP  :  ah\ 
But  the  solidity  of  a  prism  is  equal  to  the  base  multiplied  by 
the  altitude  (Prop.  XIV.)  ;  hence,  the 

prism  BCD-P  :  prism   bed-p  :  :  AH^  :  ah^  :  :  BC^  :  bc^y. 
or  as  the  cubes  of  any  other  of  their  homologous  sides. 

Cor.  Whatever  be  the  bases  of  similar  prisms,  the  prisms 
will  be  to  each  other  as  the  cubes  of  their  homologous  sides. 

For,  since  the  prisms  are  similar,  their  bases  will  be  similar 
polygons  (Def.  17.);  and  these  similar  polygons  maybe  di- 
vided into  an  equal  number  of  similar  triangles,  similarly  placed 
(Book  IV.  Prop.  XXVI.) :  therefore  the  two  prisms  may  be 
divided  into  an  equal  number  of  triangular  prisms,  having  their 
faces  similar  and  like  placed  ;  and  therefore,  equally  inclined 
(Book  VI.  Prop.  XXI.) ;  hence  the  prisms  will  be  similar.  But 
these  triangular  prisms  will  be  to  each  other  as  the  cubes  of 
their  homologous  sides,  which  sides  being  proportional,  the 
sums  of  the  triangular  prisms,  that  is,  the  polygonal  prisms,  will 
be  to  each  other  as  the  cubes  of  their  homologous  sides. 


PROPOSITION  XX.    THEOREM. 


Two  similar  pyramids  are  to  eaek  other  as  the  cubes  of  theh^ 
homologous  sides. 


BOOK  VIL  165 

For,  since  the  pyramids  are  similar,  the  soHd 
angles  at  the  vertices  will  be  contained  by  the 
same  number  of  similar  planes,  like  placed, 
and  equally  inchned  to  each  other  (Def.  17.). 
Hence,  the  solid  angles  at  the  vertices  may 
be  made  to  coincide,  or  the  two  pyramids 
may  be  so  placed  as  to  have  the  solid  angle 
S  common. 

In  that  position,  the  bases  ABCDE,  ahcde,  a/ 
will  be  parallel ;  because,  since  the  homolo- 
gous faces  are  similar,  the  angle  Safe  is  equal 
to  SAB,  and  Sfec  to  SBC ;  hence  the  plane 
ABC  is  parallel  to  the  plane  abc  (Book  VI.  Prop.  XIIL).  This 
being  proved,  let  SO  be  the  perpendicular  drawn  from  the 
vertex  S  to  the  plane  ABC,  and  o  the  point  where  this  perpen- 
dicular meets  the  plane  ahc:  from  what  has  already  been 
shown,  we  shall  have 

SO  :  So  :  :  SA  ;  Sa  :  :  AB  :    ah  (Prop.  III.) ; 
and  consequently, 

iSO  :  iSo  :  :  AB  :  ah. 
But  the  bases  ABCDE,  ahcde^  being  similar  figures,  we  have 
ABCDE  :  ahcde  :  :  AB^  ;  ah^  (Book  IV,  Prop.  XXVII.). 
Multiply  the  corresponding  terms  of  these  two  proportions ; 
there  results  the  proportion, 

ABCDE  xiSO  :  ahcde  x^So  :  :  AB^  :  ah^ 
Now  ABCDE  X  iSO  is  the  solidity  of  the  pyramid  S-ABCDE, 
and  ahcde  xjSo  is  that  of  the  pyramid  ^-ahcde  (Prop.  XVII.)  ; 
hence  two  similar  pyramids  are  to  each  other  as  the  cubes  of 
their  homologous  sides. 

General  Scholium^ 

The  chief  propositions  of  this  Book  relating  to  the  solidity  of 
polyedrons,  may  be  exhibited  in  algebraical  termsy  and  so 
recapitulated  in  the  briefest  manner  possible. 

Let  B  represent  the  base  of  a  prism ;  H  its  altitude  :  the 
solidity  of  the  prism  will  be  B  x  H,  or  BH. 

Let  B  represent  the  base  of  a  pyramid ;  H  its  altitude :  the 
solidity  of  the  pyramid  will  be  B  x  ^H,  or  H  x  ^B,  or  ^BH. 

Let  H  represent  the  altitude  of  the  frustum  of  a  pyramid, 
having  parallel  bases  A  and  B  ;  VAB  will  be  the  mean  pro- 
portional between  those  bases ;  and  the  solidity  of  the  frustum 
willbeiHx(A+B+N/AB). 

In  fine,  let  P  and  p  represent  the  solidities  of  two  similar 
prisms  or  pyramids ;  A  and  a,  two  homologous  edges  :  then  wo 
shall  have 

^  V  :  p  \  I  A?  V  a\ 


166 


GEOMETRY. 


BOOK  VIII. 


THE  THREE  ROUND  BODIES. 


Definitions, 


M 


E 


l^ 


^^ 


1.  A  cylinder  is  the  solid  generated  by  the  revolution  of  a 
rectangle  ABCD,  conceived  to  turn  about  the  immoveable 
side  AB. 

In  this  movement,  the  sides  AD,  BC,  con- 
tinuing always  perpendicular  to  AB,  describe 
equal  circles  DHP,  CGQ,  which  are  called 
the  hoses  of  the  cylinder,  the  side  CD  at  the 
same  time  describing  the  convex  surface. 

The  immoveable  line  AB  is  called  the  axis 
of  the  cylinder. 

Every  section  KLM,  made  in  the  cylinder, 
at  right  angles  to  the  axis,  is  a  circle  equal  to 
either  of  the  bases  ;  for,  whilst  the  rectangle 
ABCD  turns  about  AB,  the  line  KI,  perpen- 
dicular to  AB,  describes  a  circle,  equal  to  the  base,  and  this 
circle  is  nothing  else  than  the  section  made  perpendicular  to 
the  axis  at  the  point  I. 

Every  section  PQG,  made  through  the  axis,  is  a  rectangle 
double  of  the  generating  rectangle  ABCD. 

2.  A  cone  is  the  solid  generated  by  the  revolution  of  a  right- 
angled  triangle  SAB,  conceived  to  turn  about  the  immoveable 
side  SA. 

In  this  movement,  the  side  AB  describes 
a  circle  BDCE,  named  the  base  of  the  cone ; 
the  hypothenuse  SB  describes  the  convex 
surface  of  the  cone. 

The  point  S  is  named  the  vertex  of  the 
conCf  SA  the  axis  or  the  altitude^  and  SB 
the  side  or  the  apothem. 

Every  section  HKFI,  at  right  angles  to 
the  axis,  is  a  circle  ;  every  section  SDE, 
through  the  axis,  is  an  isosceles  triangle, 
double  of  the  generating  triangle  SAB. 

3.  If  from  the  cone  S-CDB,  the  cone  S-FKH  be  cut  off  by 
a  plane  parallel  to  the  base,  the  remaining  solid  CBHF  is  called 
a  truncated  cone,  or  the  frustum  of  a  cone. 


BOOK  VIII. 


167 


We  may  conceive  it  to  be  generated  by  the  revolution  of  a 
trapezoid  ABHG,  whose  angles  A  and  G  are  right  angles,  about 
the  side  AG.  The  immoveable  line  AG  is  called  the  axis  or 
altitude  of  the  frustum,  the  circles  BDC,  HFK,  are  its  bases,  and 
BH  is  its  side, 

4.  Two  cylinders,  or  two  cones,  are  similar,  when  their 
axes  are  to  each  other  as  the  diameters  of  their  bases. 

5.  If  in  the  circle  ACD,  which  forms  the 
base  of  a  cylinder,  a  polygon  ABODE  be 
inscribed,  a  right  prism,  constructed  on  this 
base  ABODE,  and  equal  in  altitude  to  the 
cylinder,  is  said  to  be  inscribed  in  the  cylin- 
der, or  the  cylinder  to  be  circumscribed 
about  the  prism. 

The  edges  AF,  BG,  OH,  &c.  of  the  prism, 
being  perpendicular  to  the  plane  of  the  base, 
are  evidently  included  in  the  convex  sur- 
face of  the  cylinder  ;  hence  the  prism  and 
the  cylinder  touch  one  another  along  these 
edges. 

6.  In  like  manner,  if  ABOD  is  a  poly- 
gon, circumscribed  about  the  base  of  a 
cylinder,  a  right  prism,  constructed  on  this 
base  ABOD,  and  equal  in  altitude  to  the 
cylinder,  is  said  to  be  circumsjcribed  about 
the  cylinder,  or  the  cylinder  to  be  inscribed 
in  the  prism. 

Let  M,  N,  &c.  be  the  points  of  contact 
in  the  sides  AB,  BO,  &c. ;  and  through  the 
points  M,N,&c.  let  MX,  NY,  &c.  be  drawn  Ak 
perpendicular  to  the  plane  of  the  base : 
these  perpendiculars  will  evidently  lie  both 
in  the  surface  of  the  cylinder,  and  in  that 
of  the  circumscribed  prism  ;  hence  they  will  be  their  lines  of 
contact. 

7.  If  in  the  circle  ABODE,  which  forms 
the  base  of  a  cone,  any  polygon  ABODE 
be  inscribed,  and  from  the  vertices  A,  B, 
C,  D,  E,  lines  be  drawn  to  S,  the  vertex 
of  the  cone,  these  lines  may  be  regarded 
as  the  sides  of  a  pyramid  whose  base  is 
the  polygon  ABODE  and  vertex  S.  The 
sides  of  this  pyramid  are  in  the  convex 
surface  of  the  cone,  and  the  pyramid  is 
said  to  be  inscribed  in  the  cone. 


168 


GEOMETRY. 


8.  The  sphere  is  a  solid  terminated  by  a  curved  surface,  all 
the  points  of  which  are  equally  distant  from  a  point  within, 
called  the   centre. 

The  sphere  may  be  con- 
ceived to  be  generated  by  the 
revolution  of  a  semicircle 
DAE  about  its  diameter  DE : 
for  the  surface  described  in 
this  movement,  by  the  curve 
DAE,  will  have  all  its  points 
equally  distant  from  its  cen- 
tre C. 

9.  Whilst  the  semicircle 
DAE  revolving  round  its  di- 
ameter DE,  describes  the 
sphere ;  any  circular  sector, 
as  DCF  orFCH,  describes  a 
solid,  which  is  named  a  spherical  sector. 

10.  The  radius  of  a  sphere  is  a  straight  line  drawn  from  the 
centre  to  any  point  of  the  surface  ;  the  diameter  or  axis  is  a 
line  passing  through  this  centre,  and  terminated  on  both  sides 
by  the  surface. 

All  the  radii  of  a  sphere  are  equal ;  all  the  diameters  are 
equal,  and  each  double  of  the  radius. 

11.  It  will  be  shown  (Prop.  VII.)  that  every  section  of  the 
sphere,  made  by  a  plane,  is  a  circle :  this  granted,  a  great  cir- 
cle is  a  section  which  passes  through  the  centre  ;  a  small  circle, 
is  one  which  does  not  pass  through  the  centre. 

12.  A  plane  is  tangent  to  a  sphere,  when  their  surfaces  have 
but  one  point  in  common. 

13.  A  zone  is  a  portion  of  the  surface  of  the  sphere  included 
between  two  parallel  planes,  which  form  its  bases.  One  of 
these  planes  may  be  tangent  to  the  sphere  ;  in  which  case,  the 
zone  has  only  a  single  base. 

14.  A  spherical  segment  is  the  portion  of  the  solid  sphere, 
included  between  two  parallel  planes  which  form  its  bases. 
One  of  these  planes  may  be  tangent  to  the  sphere  ;  in  which 
case,  the  segment  has  only  a  single  base. 

15.  The  altitude  of  a  zone  or  of  a  segment  is  the  distance 
between  the  two  parallel  planes,  which  form  the  bases  of  the 
zone  or  segment. 

Note.  The  Cylinder,  the  Cone,  and  the  Sphere,  are  the 
three  round  bodies  treated  of  in  the  Elements  of  Geometry. 


BOOK  VIII. 


169 


PROPOSITION  I.    THEOREM, 


The  convex  surface  of  a  cylinder  is  equal  to  the  circumference  of 
its  base  multiplied  by  its  altitude. 


Let  CA  be  the  radius  of  the 
given  cylinder's  base,  and  H  its 
ahitude  :  the  circumference 
whose  radius  is  CA  being  rep- 
resented by  circ.  CA,  we  are  to 
show  that  the  convex  surface  of 
the  cyhnder  is  equal  to  circ,  CA 

Inscribe  in  the  circle  any 
regular  polygon,  BDEFGA,  and 
construct  on  this  polygon  a  right 
prism  having  its  altitude  equal  to  H,  the  altitude  of  the  cylin- 
der :  this  prism  will  be  inscribed  in  the  cylinder.  The  convex 
surface  of  the  prism  is  equal  to  the  perimeter  of  the  polygon, 
multiplied  by  the  altitude  H  (Book  VII.  Prop.  I.).  Let  now 
the  arcs  which  subtend  the  sides  of  the  polygon  be  continually 
bisected,  and  the  number  of  sides  of  the  polygon  indefinitely 
increased  :  the  perimeter  of  the  polygon  will  then  become  equal 
to  circ.  CA  (Book  V.  Prop.  VIII.  Cor.  2.),  and  the  convex  sur- 
face of  the  prism  will  coincide  with  the  convex  surface  of  the 
cylinder.  But  the  convex  surface  of  the  prism  is  equal  to  the 
perimeter  of  its  base  multiplied  by  H,  whatever  be  the  number 
of  sides  :  hence,  the  convex  surface  of  the  cylinder  is  equal  to 
the  circumference  of  its  base  multiplied  by  its  altitude. 


PROPOSITION  II.    THEOREM. 


The  solidity  of  a  cylinder  is  equal  to  the  product  of  its  base  by  it^ 

altitude. 


170 


GEOMETRY. 


Let  CA  be  the  radius  of  the 
base  of  the  cylinder,  and  H 
the  altitude.  Let  the  circle 
whose  radius  is  CA  be  repre- 
sented by  area  CA,  it  is  to  be 
proved  that  the  solidity  of  the 
cylinder  is  equal  to  area  C  A  x  H. 
Inscribe  in  the  circle  any  regu- 
lar polygon  BDEFGA,  and  con- 
struct on  this  polygon  a  right 
prism  having  its  altitude  equal 
to  H,  the  altitude  of  the  cylinder :  this  prism  vs^ill  be  inscribed 
in  the  cylinder.  The  solidity  of  the  prism  will  be  equal  to  the 
area  of  the  polygon  multiplied  by  the  altitude  H  (Book  VIL 
Prop.  XIV.).  Let  now  the  number  of  sides  of  the  polygon  be 
indefinitely  increased  :  the  solidity  of  the  new  prism  will  still 
be  equal  to  its  base  multiplied  by  its  altitude. 

But  when  the  number  of  sides  of  the  polygon  is  indefinitely 
increased,  its  area  becomes  equal  to  the  area  CA,  and  its  pe- 
rimeter coincides  with  circ.  CA  (Book  V.  Prop.  VIII.  Cor.  1. 
&L  2.)  ;  the  inscribed  prism  then  coincides  with  the  cylinder, 
since  their  altitudes  are  equal,  and  their  convex  surfaces  per- 
pendicular to  the  common  base  :  hence  the  two  solids  will  be 
equal ;  therefore  the  solidity  of  a  cylinder  is  equal  to  the  product 
of  its  base  by  its  altitude. 

Cor.  1.  Cylinders  of  the  same  altitude  are  to  each  other  as 
their  bases  ;  and  cyHnders  of  the  same  base  are  to  each  other 
as  their  altitudes. 


Cor.  2.  Similar  cylinders  are  to  each  other  as  the  cubes  of 
their  altitudes,  or  as  the  cubes  of  the  diameters  of  their  bases. 
For  the  bases  are  as  the  squares  of  their  diameters  ;  and  the 
cylinders  being  similar,  the  diameters  of  their  bases  are  to 
each  other  as  the  altitudes  (Def  4.)  ;  hence  the  bases  are 
as  the  squares  of  the  altitudes  ;  hence  the  bases,  multiplied 
by  the  altitudes,  or  the  cylinders  themselves,  are  as  the  cubes 
of  the  altitudes. 


Scholium.  Let  R  be  the  radius  of  a  cylinder's  base  ;  H  the 
altitude  :  the  surface  of  the  base  will  be  tt.R^  (Book  V.  Prop. 
XII.  Cor.  2.)  ;  and  the  solidity  of  the  cylinder  will  be  nW  x  H, 
or^.R^.H. 


BOOK  VIII. 


171 


PROPOSITION  III.    THEOREM. 


The  convex  surf  (tee  of  a  cone  is  equal  to  the  circumference  of  its 
base,  multiplied  hy  half  its  side. 

Let  the  circle  ABCD  be  the 
base  of  a  cone,  S  the  vertex, 
SO  the  ahitude,  and  SA  the 
side  :  then  will  its  convex  sur- 
face be  equal  to  czVc.OAx  ^S  A. 

For,  inscribe  in  the  base  of 
the  cone  any  regular  polygon 
ABCD,  and  on  this  polygon  as 
a  base  conceive  a  pyramid  to 
be  constructed  having  S  for  its 
vertex :  this  pyramid  will  be  a 
regular  pyramid,  and  will  be  inscribed  in  the  cone. 

From  S,  draw  SG  perpendicular  to  one  of  the  sides  of  the 
polygon.  The  convex  surface  of  the  inscribed  pyramid  is  equal 
to  the  perimeter  of  the  polygon  which  forms  its  base,  multiplied 
by  half  the  slant  height  SG  (Book  VII.  Prop.  TV.).  Let  now 
the  number  of  sides  of  the  inscribed  polygon  be  indefinitely 
increased ;  the  perimeter  of  the  inscribed  polygon  will  then 
become  equal  to  circ.  OA,  the  slant  height  SG  will  become 
equal  to  the  side  SA  of  the  cone,  and  the  convex  surface  of 
the  pyramid  to  the  convex  surface  of  the  cone.  But  whatever 
be  the  number  of  sides  of  the  polygon  which  forms  the  base, 
the  convex  surface  of  the  pyramid  is  equal  to  the  perimeter  of 
the  base  multiplied  by  half  the  slant  height :  hence  the  convex 
surface  of  a  cone  is  equal  to  the  circumference  of  the  base 
multiphed  by  half  the  side. 

Scholium.  Let  L  be  the  side  of  a  cone,  R  the  radius  of  its 
base  ;  the  circumference  of  this  base  will  be  S^^.R,  and  the  sur- 
face of  the  cone  will  be  2nR  x  4^L,  or  nKL. 


PROPOSITION  IV.     THEOREM. 


The  convex  surface  of  the  frustum  of  a  cone  is  equal  to  its  side 
multiplied  hy  half  the  sum  of  the  circumferences  of  its  two 
bases. 


172 


GEOMETRY. 


Let  BIA-DE  be  a  frustum  of  a 
cone :  then  will  its  convex  surface  be 
equal  to  AD  x  (circ.OA+circ.CVy 

For,  inscribe  in  the  bases  of  the 
frustums  two  regular  polygons  of  the 
same  number  of  sides,  and  having 
their  homologous  sides  parallel,  each 
to  each.  The  lines  joining  the  ver- 
tices of  the  homologous  angles  may- 
be regarded  as  the  edges  of  the  frus- 
tum of  a  regular  pyramid  inscribed 
in  the  frustum  of  the  cone.  The  con- 
vex surface  of  the  frustum  of  the 
pyramid  is  equal  to  half  the  sum  of  the  perimeters  of  its  bases 
multiplied  by  the  slant  height  fh  (Book  VII.  Prop.  IV.  Cor.). 

Let  now  the  number  of  sides  of  the  inscribed  polygons  be 
indefinitely  increased :  the  perimeters  of  the  polygons  will  be- 
come equal  to  the  circumferences  BIA,  EGD  ;  the  slant  height 
^h  will  become  equal  to  the  side  AD  or  BE,  and  the  surfaces 
of  the  two  frustums  will  coincide  and  become  the  same  surface. 

But  the  convex  surface  of  the  frustum  of  the  pyramid  will 
still  be  equal  to  half  the  sum  of  the  perimeters  of  ihe  upper 
and  lower  bases  multiplied  by  the  slant  height :  hence  the  sur- 
face of  the  frustum  of  a  cone  is  equal  to  its  side  multiplied  by 
half  the  sura  of  the  circumferences  of  its  two  bases. 


Cor.  Through  /,  the  middle  point  of  AD,  draw  ZKL  paral- 
lel to  AB,  and  li,  Dd,  parallel  to  CO.  Then,  since  A/,  /D,  are 
equal,  Ai,  id,  will  also  be  equal  (Book  IV.  Prop.  XV.  Cor.  2.) : 
hence,  K/  is  equal  to  ^ (O A  +  CD).  But  since  the  circumfe- 
rences of  circles  are  to  each  other  as  their  radii  (Book  V. 
Prop.  XL),  the  circ.  K/=|(aVc.  0A  + arc.  CD) ;  therefore,  the 
convex  surface  of  a  frustum  of  a  cone  is  equal  to  its  side  multi- 
plied by  the  circumference  of  a  section  at  equal  distances  from 
the  two  bases. 

Scholium.  If  a  line  AD,  lying  wholly  on  one  side  of  the  line 
OC,  and  in  the  same  plane,  make  a  revolution  around  OC, 
the  surface  described  by  AD  will  have  for  its  measure  AD  x 
/drcAO  +  drcDCy  ^^  ^jy  ^  ^.^^  ^g- .  ^^^  jj^^^  j^^  j^^  jj^^ 

being  perpendiculars,  let  fall  from  the  extremities  and  from 
the  middle  point  of  AD,  on  the  axis  OC. 

For,  if  AD  and  OC  are  produced  till  they  meet  in  S,  the 
surface  described  by  AD  is  evidently  the  frustum  of  a  cone 


#*■. 


BOOK  Vm.  173 

having  AO  and  DC  for  the  radii  of  its  bases,  the  vertex  of 
the  whole  cone  being  S.  Hence  this  surface  will  be  measured 
as  we  have  said. 

This  measure  will  always  hold  good,  even  when  the  point 
D  falls  on  S,  and  thus  forms  a  whole  cone  ;  and  also  when  the 
hne  AD  is  parallel  to  the  axis,  and  thus  forms  a  cylinder.  In 
the  first  case  DC  would  be  nothing ;  in  the  second,  DC  would 
be  equal  to  AO  and  to  IK. 


PROPOSITION  V.    THEOREM. 

The  solidity  of  a  cone  is  equal  to  its  base  multiplied  by  a  third  of 
its  altitude. 

Let  SO  be  the  altitude  of  a  cone, 
OA  the  radius  of  its  base,  and  let 
the  area  of  the  base  be  designated 
by  area  OA :  it  is  to  be  proved  that 
the  solidity  of  the  cone  is  equal  to 
area  OAx^SO. 

Inscribe  in  the  base  of  the  cone 
any  regular  polygon  ABDEF,  and 
join  the  vertices  A,  B,  C,  &c.  with 
the  vertex  S  of  the  cone  :  then  will 
there  be  inscribed  in  the  cone  a 
regular  pyramid  having  the  same  vertex  as  the  cone,  and  hav- 
ing for  its  base  the  polygon  ABDEF.  The  solidity  of  this 
pyramid  is  equal  to  its  base  multiplied  by  one  third  of  its  alti- 
tude (Book  VII.  Prop.  XVII.).  Let  now  the  number  of  sides 
of  the  polygon  be  indefinitely  increased  :  the  polygon  will  then 
become  equal  to  the  circle,  and  the  pyramid  and  cone  will 
coincide  and  become  equal.  But  the  solidity  of  the  pyramid 
is  equal  to  its  base  multiplied  by  one  third  of  its  altitude,  what- 
ever be  the  number  of  sides  of  the  polygon  which  forms  its 
base  :  hence  the  solidity  of  the  cone  is  equal  to  its  base  multi- 
phed  by  a  third  of  its  altitude. 

Cor.  A  cone  is  the  third  of  a  cy finder  having  the  same  base 
and  the  same  altitude  ;  whence  it  follows, 

1.  That  cones  of  equal  altitudes  are  to  each  other  as  their 
bases  ; 

2.  That  cones  of  equal  bases  are  to  each  other  as  their 
altitudes ; 

3.  That  similar  cones  are  as  the  cubes  of  the  diameters  of 
their  bases,  or  as  the  cubes  of  their  altitudes. 

P* 


174  GEOMETIIY. 

Cor.  2.  The  solidity  of  a  cone  is  equivalent  to  the  solidity  of 
a  pyramid  having  an  equivalent  base  and  the  same  altitu^^ 
(Book  VII.  Prop.  XVIL). 

Scholium.  Let  R  be  the  radius  of  a  cone's  base,  H  its  alti- 
tude ;  ^he  solidity  of  the  cone  will  be  wR^  x  ^H,  or  ^nWR. 


PROPOSITION  VI.    THEOREM. 

The  solidity  of  the  frustum  of  a  cone  is  equal  to  the  sum  of  the 
solidities  of  three  cones  whose  common  altitude  is  the  altitude 
of  the  frustum,  and  whose  bases  are,  the  upper  base  of  the  frus- 
tum, the  lower  base  of  the  frustum,  and  a  mean  proportional 
between  them. 

Let  AEB-CD  be  the  frustum  of  a 
cone,  and  OP  its  altitude  ;  then  will  its 
solidity  be  equal  to 

i^r  X  OP  X  (A02+DP2+ AO  X  DP). 
For,  inscribe  in  the  lower  and  upper 
bases  two  regular  polygons  having  the 
same  number  of  sides,  and  having  their 
homologous  sides  parallel,  each  to  each. 
Join  the  vertices  of  the  homologous 
angles  and  there  will  then  be  inscribed 
in  the  frustum  of  the  cone,  the  frustum 
of  a  regular  pyramid.     The  sohdity  of 

the  frustum  of  the  pyramid  is  equivalent  to  three  pyramids 
having  the  common  altitude  of  the  frustum,  and  for  bases,  the 
lower  base  of  the  frustum,  the  upper  base  of  the  frustum,  and 
a  mean  proportional  between  them  (Book  VII.  Prop.  XVIII.). 

Let  now,  the  number  of  sides  of  the  inscribed  polygons  be 
indefinitely  increased :  the  bases  of  the  frustum  of  the  pyramid 
will  then  coincide  with  the  bases  of  the  frustum  of  the  cone, 
and  the  two  frustums  will  coincide  and  become  the  same  solid. 
Since  the  area  of  a  circle  is  equal  to  R^.n  (Book  V.  Prop.  XII. 
Cor.  2.),  the  expression  for  the  solidities  of  the  frustum  will 
become 

for  the  first  pyramid         |OP  x  OA^tt. 
for  the  second  iOP  x  PD^.tt 

for  the  third  J  OP  x  AO  x  PD.^r ;  since 

AG  X  FD.n  is  a  mean  proportional  between  OA^.^r  and  PD'.tt. 
Hence  the  solidity  of  the  frustum  of  the  cone  is  measured  by 
i^iOP  X  (OAH PDH  AO  X  PD). 


BOOK  VIII.  175 

PROPOSITION  VII.    THEOREM. 
Every  section  of  a  sphere,  made  by  a  plane,  is  a  circle. 

Let  AMB  be  a  section,  made  by  a 
plane,  in  the  sphere  whose  centre  is  C. 
From  the  point  C,  draw  CO  perpen- 
dicular to  the  plane  AMB  ;  and  diffe- 
rent lines  CM,  CM,  to  different  points 
of  the  curve  AMB,  which  terminates 
the  section. 

The  oblique  lines  CM,  CM,  CA,  are 
equal,  being  radii  of  the  sphere  ;  hence 
they  are  equally  distant  from  the  perpendicular  CO  (Book  VL 
Prop.  V.  Cor.)  ;  therefore  all  the  lines  OM,  OM,  OB,  are  equal ; 
consequently  the  section  AMB  is  a  circle,  whose  centre  is  Q. 

Cor  1.  If  the  section  passes  through  the  centre  of  the  sphere, 
its  radius  will  be  the  radius  of  the  sphere ;  hence  all  great 
circles  are  equal. 

Cor.  2.  Two  great  circles  always  bisect  each  other ;  for 
their  common  intersection,  passing  through  the  centre,  is  a 
diameter. 

Cor.  3.  Every  great  circle  divides  the  sphere  and  its  surface 
into  two  equal  parts :  for,  if  the  two  hemispheres  were  sepa- 
rated and  afterwards  placed  on  the  common  base,  with  their 
convexities  turned  the  same  way,  the  two  surfaces  would 
exactly  coincide,  no  point  of  the  one  being  nearer  the  centre 
than  any  point  of  the  other. 

Cor.  4.  The  centre  of  a  small  circle,  and  that  of  the  sphere, 
are  in  the  same  straight  line,  perpendicular  to  the  plane  of  the 
small  circle. 

Cor.  5.  Small  circles  are  the  less  the  further  they  lie  from 
the  centre  of  the  sphere  ;  for  the  greater  CO  is,  the  less  is  the 
chord  AB,  the  diameter  of  the  small  circle  AMB. 

Cor.  6.  An  arc  of  a  great  circle  may  always  be  made  to  pass 
through  any  two  given  points  of  the  surface  of  the  sphere  ;  for 
the  two  given  points,  and  the  centre  of  the  sphere  make  three 
points  which  determine  the  position  of  a  plane.  But  if  the 
two  given  points  were  at  the  extremities  of  a  diameter,  these 
two  points  and  the  centre  would  then  lie  in  one  straight  line, 
and  an  infinite  number  of  great  circles  might  be  made  to  pass 
through  the  two  given  points. 


176 


GEOMETRY. 


PROPOSITION  VIII.    THEOREM. 

Evety  plane  perpendiculaj-  to  a  radius  at  its  extremity  is  tangent 
to  the  sphere. 

Let  FAG  be  a  plane  perpendicular 
to  the  radius  OA,  at  its  extremity  A. 
Any  point  M  in  this  plane  being  as- 
sumed, and  OM,  AM,  being  drawn, 
the  angle  0AM  will  be  a  right  angle, 
and  hence  the  distance  OM  will  be 
greater  than  OA.  Hence  the  point 
M  lies  without  the  sphere  ;  and  as  the 
same  can  be  shown  for  every  other 
point  of  the  plane  FAG,  this  plane  can 
have  no  point  but  A  common  to  it  and  the  surface  of  the  sphere ; 
hence  it  is  a  tangent  plane  (Def.  12.) 

Scholium.  In  the  same  way  it  may  be  shown,  that  two 
spheres  have  but  one  point  in  common,  and  therefore  touch 
each  other,  when  the  distance  between  their  centres  is  equal  to 
the  sum,  or  the  difference  of  their  radii ;  in  which  case,  the 
centres  and  the  point  of  contact  lie  in  the  same  straight  line. 


PROPOSITION  IX.    LEMMA. 


If  a  regular  semi-polygon  he  revolved  about  a  line  passing 
through  the  centre  and  the  vertices  of  two  opposite  angles,  the 
surface  described  by  its  perimeter  will  be  equal  to  the  axis  mul- 
tiplied by  the  circumference  of  the  inscribed  circle. 

Let  the  regular  semi-polygon  ABCDEF, 
be  revolved  about  the  line  AF  as  an  axis  : 
then  will  the  surface  described  by  its  pe- 
rimeter be  equal  to  AF  multiplied  by  the 
circumference  of  the  inscribed  circle. 

From  E  and  D,  the  extremities  of  one  of 
the  equal  sides,  let  fall  the  perpendiculars 
EH,  DI,  on  the  axis  AF,  and  from  the  cen- 
tre O  draw  ON  perpendicular  to  the  side 
DE :  ON  will  be  the  radius  of  the  inscribed 
circle  (Book  V.  Prop.  XL).  Now,  the  sur- 
face described  in  the  revolution  by  any  one 
side  of  the  regular  polygon,  as  DE,  has 


BOOK  VIII. 


177 


been  shown  to  be  equal  to  DE  x  circ.  NM  (Prop.  IV.  Sch.). 
But  since  the  triangles  EDK,  ONM,  are  similar  (Book  IV. 
Prop.  XXL),  ED  ;  EK  or  HI  :  :  ON  :  NM,  or  as  circ.  ON  ; 
circ.  NM ;  hence 

ED  X  circ.  NM=HI  X  circ.  ON ; 
and  since  the  same  may  be  shown  for  each  of  the  other  sides, 
it  is  plain  that  the  surface  described  by  the  entire  perimeter  is 
equal  to 

(FH  +  HH-  IP + PQ  +  QA)  X  circ  0N= AF  x  circ.  ON. 

Cor.  The  surface  described  by  any  portion  of  the  perime- 
ter, as  EDC,  is  equal  to  the  distance  between  the  two  perpen- 
diculars let  fall  from  its  extremities  on  the  axis,  multiplied  by 
the  circumference  of  the  inscribed  circle.  For,  the  surface 
described  by  DE  is  equal  to  HI  x  circ.  ON,  and  the  surface 
described  by  DC  is  equal  to  IP  x  cii'c.  ON  ;  hence  the  surface 
described  by  ED  +  DC,  is  equal  to  (HI  +  IP)  x  circ.  ON,  or 
equal  to  HP  x  circ.  ON. 


PROPOSITION  X.     THEOREM. 


The  surface  of  a  sphere  is  equal  to  the  product  of  its  diameter  hy 
the  circumference  of  a  great  circle. 

Let  ABCDE  be  a  semicircle.  Inscribe  in 
it  any  regular  semi-polygon,  and  from  the 
centre  O  draw  OF  perpendicular  to  one  of 
the  sides. 

Let  the  semicircle  and  the  semi-polygon 
be  revolved  about  the  axis  AE  :  the  semi- 
circumference  ABCDE  will  describe  the 
surface  of  a  sphere  (Def  8.) ;  and  the  pe- 
rimeter of  the  semi-polygon  will  describe 
a  surface  which  has  for  its  measure  AE  x 
circ.  OF  (Prop.  IX.),  and  this  will  be  true 
whatever  be  the  number  of  sides  of  the  po- 
lygon. But  if  the  number  of  sides  of  the  polygon  be  indefi- 
nitely increased,  its  perimeter  will  coincide  with  the  circumfe- 
rence ABCDE,  the  perpendicular  OF  will  become  equal  to 
OE,  and  the  surface  described  by  the  perimeter  of  the  semi- 
polygon  will  then  be  the  same  as  that  described  by  the  semi- 
circumference  ABCDE.  Hence  the  surface  of  the  sphere  is 
equal  to  AE  x  circ.  OE. 

Cor.  Since  the  area  of  a  great  circle  is  equal  to  the  product 
\     of  its  circumference  by  half  the  radius,  or  one  fourth  of  the 

23 


178 


GEOMETRY. 


V 


diameter  (Book  V.  Prop.  XII.),  it  follows  that  the  surface  of  a 
sphere  is  equal  to  four  of  its  great  circles :  that  is,  equal  to 
4^.0 A2  (Book  V.  Prop.  XII.  Cor.  2.). 

Scholium  1.  The  surface  of  a  zone  is  equal  to  its  altitude  mul- 
tiplied by  the  circumference  of  a  great  circle. 

For,  the  surface  described  by  any  portion 
of  the  perimeter  of  the  inscribed  polygon,  as 
BC  +  CD,  is  equal  to  EH  xcirc.  OF  (Prop. 
IX.  Cor.).  But  when  the  number  of  sides 
of  the  polygon  is  indefinitely  increased,  BC 
+  CD,  becomes  the  arc  BCD,  OF  becomes 
equal  to  OA,  and  the  surface  described  by 
BC  +  CD,  becomes  the  surface  of  the  zone 
described  by  the  arc  BCD;  hence  the  sur- 
face of  the  zone  is  equal  to  EH  x  circ.  OA. 

Scholium  2.  When  the  zone  has  but  one 
base,  as  the  zone  described  by  the  arc  ABC D,  its  surface  will 
still  be  equal  to  the  altitude  AE  multiplied  by  the  circumference 
of  a  great  circle. 

Scholium  3.  Two  zones,  taken  in  the  same  sphere  or  in  equal 
spheres,  are  to  each  other  as  their  altitudes  ;  and  any  zone  is  to 
the  surface  of  the  sphere  as  the  altitude  of  the  zone  is  to  the 
diameter  of  the  sphere. 


PKGPOSITION  XL    LEMMA. 


If  a  triangle  and  a  rectangle ^  having  the  same  base  and  the  same 
altitudCfturn  together  about  the  common  base^  the  solid  described 
by  the  triangle  will  be  a  third  of  the  cylinder  described  by  the 
rectangle. 

Let  ACB  be  the  triangle,  and  BE  the  rectangle. 

On  the  axis,  let  fall  the  perpen- 
dicular AD :  the  cone  described  by 
the  triangle  ABD  is  the  third  part  of 
the  cylinder  described  by  the  rectan- 
gle AFBD  (Prop.  V.  Cor.) ;  also  the 
cone  described  by  the  triangle  ADC 
is  the  third  part  of  the  cylinder  de- 
scribed by  the  rectangle  ADCE  ;  hence  the  sum  of  the  two 
cones,  or  the  solid  described  by  ABC,  is  the  third  part  of  the 
two  cylinders  taken  together,  or  of  the  cylinder  described  by 
the  rectangle  BCEF. 


BOOK  VIIL 


17^ 


If  the  perpendicular  AD  falls  without 
the  triangle ;  the  solid  described  by  ABC 
will,  in  that  case,  be  the  difference  of  the 
two  cones  described  by  ABD  and  ACD  ; 
but  at  the  same  time,  the  cylinder  de- 
scribed by  BCEF  will  be  the  difference  ^  ^  ^ 
of  the  two  cylinders  described  by  AFBD  and  AECD.  Hence 
the  solid,  described  by  the  revolution  of  the  triangle,  will  still 
be  a  third  part  of  the  cylinder  described  by  the  revolution  of 
the  rectangle  having  the  same  base  and  the  same  altitude. 

Scholium.  The  circle  of  which  AD  is  radius,  has  for  its 
measure  n  x  AD^ ;  hence  n  x  AD^  x  BC  measures  the  cylinder 
described  by  BCEF,  and  ^n  x  AD^  x  BC  measures  the  solid 
described  by  the  triangle  ABC. 


PROPOSITION  XII.    LEMMA. 


If  a  triangle  he  revolved  about  a  line  drawn  at  pleasure  through 
its  vertex,  the  solid  described  by  the  triangle  will  have  for  its 
measure^  the  area  of  the  triangle  multiplied  by  two  thirds  of  the 
circumference  traced  by  the  middle  point  of  the  base. 

Let  CAB  be  the  triangle,  and  CD  the  line  about  which  it 
revolves. 

Produce  the  side  AB  till  it 
meets  the  axis  CD  in  D  ;  from  the 
points  A  and  B,  draw  AM,  BN, 
perpendicular  to  the  axis,  and  CP 
perpendicular  to  DA  produced. 

The  solid  described  by  the  tri- 
angle CAD  is  measured  by  ^n  x 
AM^  X  CD  (Prop.  XI.  Sch.) ;  the  solid  described  by  the  triangle 
CBD  is  measured  by  ^n  x  BN^  x  CD  ;  hence  the  difference  of 
those  solids,  or  the  solid  described  by  ABC,  will  have  for  its 
measure  iTr(AM2— BN^)  x  CD. 

To  this  expression  another  form  may  be  given.  From  I,  the 
middle  point  of  AB,  draw  IK  perpendicular  to  CD  ;  and  through 
B,  draw  BO  parallel  to  CD:  we  shall  have  AM  +  BN=2IK 
(Book  IV.  Prop.  VII.) ;  and  AM— BN=AO  ;  hence  (AM  + 
BN)  X  (AM— NB),  or  AM^— BN2=2lKx  AO  (Book  IV.  Prop. 
X.).  Hence  the  measure  of  the  solid  in  question  is  ex- 
pressed by 

f^xIKxAOxCD. 


GEOMETRY. 


JVLKJSr 


But  CP  being  drawn  perpendicular  to  AB,  the  triangles  ABO, 
DCP  will  be  similar,  and  give  the  proportion 

AO  :  CP  :  :  AB  :  CD  ; 
hence  AO  x  CD = CP  x  AB  ; 

but  CP  X  AB  is  double  the  area  of  the  triangle  ABC  ;  hence 
we  have 

A0xCD=2ABC; 

hence  the  solid  described  by  the 

triangle  ABC  is  also  measured 

by  |7r  X  ABC  X  IK,  or  which  is  the 

same  thing,  by  ABC  x  Icirc.  IK, 

circ.  IK  being  equal  to  27i^  x  IK. 

Hence  the  solid  described  hy  the 

revolution  of  the  triangle  A^C, has 

for  its  measure  the  area  of  this  triangle  multiplied  by  two  thirds 

of  the  circumference  traced  by  I,  the  middle  point  of  the  base. 

Cor.  If  the  sideAC  =  CB, 
the  line  CI  will  be  perpen- 
dicular to  AB,  the  area  ABC 
will  be  equal  to  AB  x  |CI, 
and  the  solidity  ^71  x  ABC  x 
IK  will  become  fTtxABx 
IKxCI.  But  the  triangles 
ABO,  CIK,  are  similar,  and 
give  the  proportion  AB  :  BO 
or  MN  :  ;  CI  :  IK;  hence  ABxIK=MNxCI;  hence  the 
solid  described  by  the  isosceles  triangle  ABC  will  have  for  its 
measure  ^n  x  CP  x  MN. 

Scholium.  The  general  solution  appears  to  include  the  sup- 
position that  AB  produced  will  meet  the  axis  ;  but  the  results 
would  be  equally  true,  though  AB  were  parallel  to  the  axis. 

Thus,  the  cylinder  described  by  AMNB   p  i\  B 

is  equal  to  jt.AM^.MN  ;  the  cone  descri- 
bed by  ACM  is  equal  to  i^r.AM^.CM, 
and  the  cone  described  by  BCN  to 
inAM^  CN.  Add  the  first  two  sohds  and 
take  away  the  third  ;  we  shall  have  the 
solid  described  by  ABC  equal  to  tt.AM^ 
(MN  +  iCM— iCN):  and  since  CN— CM =MN,  this  expres- 
sion is  reducible  to  tt.AM^.^MN,  or  Iti.CP^.MN  ;  which  agrees 
with  the  conclusion  found  above. 


BOOK  VIIL 


181 


PROPOSITION  XIII.    LEMMA. 


If  a  regular  semi-polygon  he  revolved  about  a  line  passing 
through  the  centre  and  the  vertices  of  two  opposite  angles,  the 
solid  described  will  be  equivalent  to  a  cone,  having  for  its  base 
the  inscribed  circle,  and  for  its  altitude  twice  the  axis  about 
which  the  semi-polygon  is  revolved. 


Let  the  semi-polygon  FABG  be  revolved 
about  FG  :  then,  if  OI  be  the  radius  of  the 
inscribed  circle,  the  solid  described  will  be 
measured  by  ^area  01  x  2FG. 

For,  since  the  polygon  is  regular,  the 
triangles  OFA,  OAB,  OBC,  &c.  are  equal 
and  isosceles,  and  all  the  perpendiculars  let 
fall  from  O  on  the  bases  FA,  AB,  (fee.  will 
be  equal  to  01,  the  radius  of  the  inscribed 
circle. 

Now,  the  solid  described  by  OAB  has 
forits  measure  |0P  X  MN  (Prop.  Xll.Cor.); 
the  solid  described  by  the  triangle  OFA  has  for  its  measure 
fTiOP  X  FM,  the  solid  described  by  the  triangle  OBC,  has  for 
its  measure  f  tiOP  x  NO,  and  since  the  same  may  be  shown  for 
the  solid  described  by  each  of  the  other  triangles,  it  Ibllows 
that  the  entire  solid  described  by  the  semi-polygon  is  mea- 
sured by  |nOP.(FM  +  MN  +  NO  +  OQ  +  QG),or  |7>0FxFG; 
which  is  also  equal  to  \nO\^  x  2FG.  But  yr.OP  is  the  area  of 
the  inscribed  circle  (Book  V.  Prop,  XII.  Cor.  2.) :  hence  the 
solidity  is  equivalent  to  a  cone  whose  base  is  area  01,  and 
altitude  2FG. 


PROPOSITION  XIV.    THEOREM. 

The  solidity  of  a  sphere  is  equal  to  its  surface  multiplied  by  a 
third  of  its  radius. 


a 


182  GEOMETRIC. 

Inscribe  in  the  semicircle  ABCDE  a 
regular  semi-polygon,  having  any  number 
of  sides,  and  let  01  be  the  radius  of  the 
circle  inscribed  in  the  polygon. 

If  the  semicircle  and  semi-polygon  be 
revolved  about  EA,  the  semicircle  will 
describe  a  sphere,  and  the  semi-polygon  a 
solid  which  has  for  its  measure  f^OP  x 
EA  (Prop.  XIII.) ;  and  this  will  be  true 
whatever  be  the  number  of  sides  of  the 
polygon.  But  if  the  number  of  sides  of 
the  polygon  be  indefinitely  increased,  the  E 

semi-polygon  will  become  the  semicircle,  01  will  become 
equal  to  OA,  and  the  solid  described  by  the  semi-polygon  will 
become  the  sphere  :  hence  the  solidity  of  the  sphere  is  equal 
to  fTiOA^xEA,  or  by  substituting  20A  for  EA,  it  becomes 
jn.OA^  X  OA,  which  is  also  equal  to  4nOA^  x  ^OA.  But  4n.OA^ 
is  equal  to  the  surface  of  the  sphere  (Prop.  X.  Cor.) :  hence 
the  solidity  of  a  sphere  is  equal  to  its  surface  multiplied  by  a 
third  of  its  radius. 

Scholium  1.  The  solidity  of  every  spherical  sector  is  equal  to 
the  zone  which  forms  its  base,  multiplied  by  a  third  of  the  radius. 

For,  the  solid  described  by  any  portion  of  the  regular  poly- 
gon, as  the  isosceles  triangle  OAB,  is  measured  by  fyrOP  x  AF 
(Prop.  XII.  Cor.) ;  and  when  the  polygon  becomes  the  circle, 
the  portion  OAB  becomes  the  sector  AOB,  01  becomes  equal 
to  OA,  and  the  solid  described  becomes  a  spherical  sector.  But 
its  measure  then  becomes  equal  to  fyr. AO^  x  AF,  which  is  equal 
to  27r. AO  X  AF  X  |AO.  But  27r.AO  is  the  circumference  of  a 
great  circle  of  the  sphere  (Book  V.  Prop.  XII.  Cor.  2.),  which 
being  multiplied  by  AF  gives  the  surface  of  the  zone  which 
forms  the  base  of  the  sector  (Prop.  X.  Sch.  1.) :  and  the  proof 
is  equally  applicable  to  the  spherical  sector  described  by  the 
circular  sector  BOC  :  hence,  the  solidity  of  the  spherical  sector 
is  equal  to  the  zone  which  forms  its  base,  multiplied  by  a  third 
of  the  radius. 

Scholium  2.  Since  the  surface  of  a  sphere  whose  radius  is 
R,  is  expressed  by  47rR2  (Prop.  X.  Cor.),  it  follows  that  the 
surfaces  of  spheres  are  to  each  other  as  the  squares  of  their 
radii ;  and  since  their  solidities  are  as  their  surfaces  multiplied 
by  their  radii,  it  follows  that  the  solidities  of  spheres  are  to 
each  other  as  the  cubes  of  their  radii,  or  as  the  cubes  of  their 
diameters. 


BOOK  VIII. 


183 


Scholium  3.  Let  R  be  the  radius  of  a  sphere  ;  its  surface 
will  be  expressed  by  47iR^,  and  its  solidity  by  47iR2  x  ^R,  or 
^TiR^.  If  the  diameter  is  called  D,  we  shall  have  R=|D, 
and  R^=|D^  :  hence  the  solidity  of  the  sphere  may  likewise  be 
expressed  by 


PROPOSITION  XV.    THEOREM. 


Tfie  surface  of  a  sphere  is  to  the  whole  surface  of  the  circum- 
scribed cylinder,  including  its  bases,  as  2  is  to  3  ;  and  the  so- 
lidities of  these  two  bodies  are  to  each  other  in  the  same  ratio. 


D 


Let  M PNQ  be  a  great  circle  of  the 
sphere ;  ABCD  the  circumscribed 
square  :  if  the  semicircle  PMQ  and 
the  half  square  PADQ  are  at  the 
same  time  made  to  revolve  about  the 
diameter  PQ,  the  semicircle  will  gene-  M 
rate  the  sphere,  while  the  half  square 
will  generate  the  cylinder  circum- 
scribed about  that  sphere.  . 

The  altitude  AD  of  the  cylinder  is 
equal  to  the  diameter  PQ  ;  the  base  of 
the  cylinder  is  equal  to  the  great  circle,  since  its  diameter  AB 
is  equal  to  MN  ;  hence,  the  convex  surface  of  the  cylinder  is 
equal  to  the  circumference  of  the  great  circle  multiplied  by  its 
diameter  (Prop.  1.).  This  measure  is  the  same  as  that  of  the 
surface  of  the  sphere  (Prop.  X.) ;  hence  the  surface  of  the  sphere 
is  equal  to  the  convex  surface  of  the  circumscribed  cylinder. 

But  the  surface  of  the  sphere  is  equal  to  four  great  circles  ; 
hence  the  convex  surface  of  the  cylinder  is  also  equal  to  four 
great  circles  :  and  adding  the  two  bases,  each  equal  to  a  great 
circle,  the  total  surface  of  the  circumscribed  cylinder  will  be 
equal  to  six  great  circles ;  hence  the  surface  of  the  sphere  is  to 
the  total  surface  of  the  circumscribed  cylinder  as  4  is  to  6,  or 
as  2  is  to  3  ;  which  was  the  first  branch  of  the  Proposition. 

In  the  next  place,  since  the  base  of  the  circumscribed  cylin- 
der is  equal  to  a  great  circle,  and  its  altitude  to  the  diameter, 
the  solidity  of  the  cylinder  will  be  equal  to  a  great  circle  mul- 
tiplied by  its  diameter  (Prop.  II.).  But  the  solidity  of  the 
sphere  is  equal  to  four  great  circles  multiplied  by  a  third  of  the 
radius  (Prop.  XIV.);  in  other  terms,  to  one  great  circle  multi- 
plied by  I  of  the  radius,  or  by  f  of  the  diameter ;  hence  the 
sphere  is  to  the  circumscribed  cylinder  as  2  to  3,  and  conse- 
quently the  solidities  of  these  two  bodies  are  as  their  surfaces. 


184  GEOMETRY.  * -^ 

Scholium.  Conceive  a  polyedron,  all  of  whose  faces  touch 
the  sphere  ;  this  polyedron  may  be  considered  as  formed  of 
pyramids,  each  having  for  its  vertex  the  centre  of  the  sphere, 
and  for  its  base  one  of  the  polyedron's  faces.  Now  it  is  evi- 
dent that  all  these  pyramids  will  have  the  radius  of  the  sphere 
for  their  common  altitude :  so  that  each  pyramid  will  be  equal 
to  one  face  of  the  polyedron  multiplied  by  a  third  of  the  radius  : 
hence  the  whole  polyedron  will  be  equal  to  its  surface  multi- 
plied by  a  third  of  the  radius  of  the  inscribed  sphere. 

It  is  therefore  manifest,  that  the  solidities  of  polyedrons  cir- 
cumscribed about  the  sphere  are  to  each  other  as  the  surfaces 
of  those  polyedrons.  Thus  the  property,  which  we  have  shown 
to  be  true  with  regard  to  the  circumscribed  cylinder,  is  also 
true  with  regard  to  an  infinite  number  of  other  bodies. 

We  might  likewise  have  observed  that  the  surfaces  of  poly- 
gons, circumscribed  about  the  circle,  are  to  each  other  as  their 
perimeters. 


PROPOSITION  XVI.    PROBLEM. 

If  a  circular  segment  he  supposed  to  make  a  revolution  about  a 
diameter  exterior  to  it,  required  the  value  of  the  solid  which  it 
describes. 

Let  the  segment  BMD  revolve  about  AC. 

On  the  axis,  let  fall  the  perpendiculars 
BE,  DF  ;  from  the  centre  C,  draw  CI 
perpendicular  to  the  chord  BD  ;  also  draw 
the  radii  CB,  CD. 

The  solid  described  by  the  sector  BCA 
is  equal  to  fTr.CB^.AE  (Prop.  XIV.  Sch. 
1.) ;  the  solid  described  by  the  sector 
DCA=|7r.CB2.AF;  hence  the  difference 
of  these  two  solids,  or  the  solid  described 
by  the  sector  DCB=|^.CB2.(AF— AE)  =  |rr.CB2.EF.  But  the 
solid  described  by  the  isosceles  triangle  DCB  has  for  its  mea- 
sure f^.CP.EF  (Prop.  XII.  Cor.) ;  hence  the  solid  described 
by  the  segment  BMD=|rr.EF.(CB^— CP).  Now,  in  the  right- 
angled  triangle  CBI,  we  have  CB^—CP^BP^^BD^  ;  hence 
the  solid  described  by  the  segment  BMD  will  have  for  its  mea- 
sure |7..EF.iBD2,  or  i^.BD-.EF. 

Scholium.  The  solid  described  by  the  segment  BMD  is  to 
the  sphere  which  has  BD  for  its  diameter,  as  i7T.BD-.EF  is 
to  i^.BD^  or  as  EF  to  BD, 


BOOK  VIII. 


185 


PROPOSITION  XVII.    THEOREM. 

Every  segment  of  a  sphere^  included  between  two  parallel  planes^ 
is  measured  by  the  half  sum  of  its  bases  multiplied  by  its  alti- 
tudet  plus  the  solidity  of  a  sphere  whose  diameter  is  this  same 
altitude. 

Let  BE,  DF,  be  the  radii  of  the  two 
bases  of  the  segment,  EF  its  altitude,  the 
segment  being  described  by  the  revolu- 
tion of  the  circular  space  BMDFE  about 
the  axis  FE.  The  solid  described  by  the 
segment  BMD  is  equal  to  ^Tr.BD^.EF 
(Prop.  XVI.) ;  and  the  truncated  cone  de- 
scribed by  the  trapezoid  BDFE  is  equal 
toi7r.EF.(BEHDF-+BE.DF)(Prop.VI.); 
hence  the  segment  of  the  sphere,  which  is  the  sum  of  those  two 
solids,  must  be  equal  to  i^.EF.(2BEH2DF2H-2BE.DF+BD?) 
But,  drawing  BO  parallel  to  EF,  we  shall  have  DO=DF— BE, 
hence  DO^^DF— 2DF.BE  +  BE«  (Book  IV.  Prop.  IX.)  ;  and 
consequently  BD^^BOH  D02=:EF2+ DF— 2DF .  BE  +  BEl 
Put  this  value  in  place  of  BD^  in  the  expression  for  the  value 
of  the  segment,  omitting  the  parts  which  destroy  each  other ; 
we  shall  obtain  for  the  solidity  of  the  segment, 

i7rEF.(3BE^+3DF2+EF), 
an  expression  which  may  be  decomposed  into  two  parts  ;  the 

/TT.BEH^.DFx 
one  i7r.EF.(3BEH3DF2),  or  EF.( ^ )  being  the 

half  sum  of  the  bases  multiplied  by  the  altitude  ;  while  the 
other  iTT.EF^  represents  the  sphere  of  which  EF  is  the  diame- 
ter (Prop.  XIV.  Sch.) :  hence  every  segment  of  a  sphere,  &c. 

Cor.  If  either  of  the  bases  is  nothing,  the  segment  in  ques- 
tion becomes  a  spherical  segment  with  a  single  base  ;  hence 
any  spherical  segment^  with  a  single  base,  is  equivalent  to  half 
the  cylinder  having  the  same  base  and  the  same  altitude,  plus  the 
sphere  of  which  this  altitude  is  the  diameter. 


General  Scholium. 

Let  R  be  the  radius  of  a  cylinder's  base,  H  its  altitude  ;  the 
solidity  of  the  cylinder  will  be  nR^  x  H,  or  tiR^H. 

Let  R  be  the  radius  of  a  cone's  base,  H  its  altitude :  the 
solidity  of  the  cone  will  be  nR^  x  ^H,  or  ^tiR^H. 

Let  A  and  B  be  the  radii  of  the  bases  of  a  truncated  cone, 

a*  24 


186  GEOMETRY. 

H  its  altitude  :  the  solidity  of  the  truncated  cone  will  be  |7r.H. 
(AHB2+AB). 

Let  R  be  the  radius  of  a  sphere  ;  its  solidity  will  be  ^ttR^. 

Let  R  be  the  radius  of  a  spherical  sector,  H  the  altitude  of 
the  zone,  which  forms  its  base  :  the  solidity  of  the  sector  will 
be  f  ^R2H. 

Let  P  and  Q  be  the  two  bases  of  a  spherical  segment,  H  its 

piQ 

altitude:  the  solidity  of  the  segment  will  be         ^.H+^rr.H^ 

til 

If  the  spherical  segment  has  but  one  base,  the  other  being 
nothing,  its  solidity  will  be  iPH+^TiH^. 

BOOK  IX. 

OP  SPHERICAL  TRIANGLES  AND  SPHERICAL  POLYGONS. 


Definitions, 

1.  A  spherical  triangle  is  a  portion  of  the  surface  of  a  sphere, 
bounded  by  three  arcs  of  great  circles. 

These  arcs  are  named  the  sides  of  the  triangle,  and  are 
always  supposed  to  be  each  less  than  a  semi-circumference. 
The  angles,  which  their  planes  form  with  each  other,  are  the 
angles  of  the  triangle. 

2.  A  spherical  triangle  takes  the  name  of  right-angled, 
isosceles,  equilateral,  in  the  same  cases  as  a  rectilineal  triangle. 

3.  A  spherical  polygon  is  a  portion  of  the  surface  of  a  sphere 
terminated  by  several  arcs  of  great  circles. 

4.  A  lune  is  that  portion  of  the  surface  of  a  sphere,  which  is 
included  between  two  great  semi-circles  meeting  in  a  common 
.diameter. 

5.  A  spherical  wedge  or  ungula  is  that  portion  of  the  solid 
sphere,  which  is  included  between  the  same  great  semi-circles, 
and  has  the  lune  for  its  base. 

6.  A  spherical  pyramid  is  a  portion  of  the  solid  sphere,  in- 
cluded between  the  planes  of  a  solid  angle  whose  vertex  is 
the  centre.  The  base  of  the  pyramid  is  the  spherical  polygon 
intercepted  by  the  same  planes. 

7.  The  pole  of  a  circle  of  a  sphere  is  a  point  in  the  surface 
equally  distant  from  all  the  points  in  the  circumference  of  this 
circle.     It  will  be  shown  (Prop.  V.)  that  every  circle,  great  or 

1^  small,  has  always  two  poles. 


BOOK  IX.  187 


PROPOSITION  I.    THEOREM. 

In  every  spherical  triangle^  any  side  is  less  than  the  sum  of  the 

other  two. 

Let  O  be  the  centre  of  the  sphere,  and 
ACB  the  triangle  ;  draw  the  radii  OA,  OB, 
OC.  Imagine  the  planes  AOB,  AOC, 
COB,  to  be  drawn  ;  these  planes  will  form 
a  solid  angle  at  the  centre  O  ;  and  the  an- 
gles AOB,  AOC,  COB,  will  be  measured" 
by  AB,  AC,  BC,  the  sides  of  the  spherical 
triangle.  But  each  of  the  three  plane  an- 
gles forming  a  solid  angle  is  less  than  the 
sum  of  the  other  two  (Book  VI.  Prop. 
XIX.) ;  hence  any  side  of  the  triangle 
ABC  is  less  than  the  sum  of  the  other  two. 

PROPOSITION  II.    THEOREM. 

The  shortest  path  from  one  point  to  another,  on  the  surface  of  a 
sphere,  is  the  arc  of  the  great  circle  which  joins  the  two  given 
points. 

Let  ANB  be  the  arc  of  the  great  circle  which  joins 
the  points  A  and  B  ;  and  without  this  hne,  if  possible, 
let  M  be  a  point  of  a  shorter  path  between  A  and 
B.  Through  the  point  M,  draw  MA,  MB,  arcs  of 
great  circles  ;  and  take  BN=MB. 

By  the  last  theorem,  the  arc  ANB  is  shorter  than 
AM  +  MB  ;  take  BN^BM,  from  both  ;  there  will  re- 
main  AN<AM.  Now,  the  distance  of  B  from  M,  -^ 
whether  it  be  the  same  with  the  arc  BM  or  with  any 
other  line,  is  equal  to  the  distance  of  B  from  N ;  for 
by  making  the  great  circle  BM  to  revolve  about  the 
diameter  which  passes  through  B,  the  point  M  may  ^ 
be  brought  into  the  position  of  the  point  N  ;  and  the 
shortest  line  between  M  and  B,  whatever  it  may  be,  will  then 
be  identical  with  that  between  N  and  B :  hence  the  two  paths 
from  A  to  B,  one  passing  through  M,  the  other  through  N,  have 
an  equal  part  in  each,  the  part  from  M  to  B  equal  to  the  part 
from  N  to  B.  The  first  path  is  the  shorter  by  hypothesis ; 
hence  the  distance  from  A  to  M  must  be  shorter  than  the  dis- 
tance from  A  to  N  ;  which  is  absurd,  the  arc  AM  being  proved 
greater  than  AN ;  hence  no  point  of  the  shortest  line  from  A 
to  B  can  lie  out  of  the  arc  ANB ;  hence  this  arc  is  itself  the 
shortest  distance  between  its  two  extremities.* 

*  Note.  This  demonstration  of  Legendre  appears  to  be  inconclusive.— Ed. 


■IS- 


188  GEOMETRY. 


PROPOSITION  III.    THEOREM. 


The  sum  of  the  three  sides  of  a  spherical  triangle  is  less  than  the 
circumference  of  a  great  circle. 

Let  ABC  be  any  spherical  trian- 
angle  ;  produce  the  sides  AB,  AC, 
till  they  meet  again  in  D.  The 
arcs  ABD,  ACD,  will  be  semicir- 
cumferences,  since  two  great  cir- 
cles always  bisect  each  other  (Book 
VIII.  Prop.  VII.  Cor.  2.).  But  in 
the  triangle  BCD,  we  have  the  side 
BC<BD  +  CD  (Prop.  I.) ;  add  AB 
-f  AC  to  both  ;  we  shall  have  AB-f 
AC+BC<ABD+ACD,  thatis  to 
say,  less  than  a  circumference. 


PROPOSITION  IV.    THEOREM. 

The  sum  of  all  the  sides  of  any  spherical  polygon  is  less  than  the 
circumference  of  a  great  circle. 

Take  the  pentagon  ABCDE,  for 
example.  Produce  the  sides  AB,  DC, 
till  they  meet  in  F ;  then  since  BC  is 
less  than  BF  +  CF,  the  perimeter  of 
the  pentagon  ABCDE  will  be  less 
than  that  of  the  quadrilateral  AEDF. 
Again,  produce  the  sides  AE,  FD,  till 
they  meet  in  G ;  we  shall  have  ED<EG+DG ;  hence  the  pe- 
rimeter of  the  quadrilateral  AEDF  is  less  than  that  of  the  tri- 
angle AFG;  which  last  is  itself  less  than  the  circumference  of 
a  great  circle  ;  hence,  for  a  still  stronger  reason,  the  perimeter 
of  the  polygon  ABCDE  is  less  than  this  same  circumference. 


^*^^ipiV^, 


BOOK  IX. 


189 


Scholium.  This  proposition  is  fun- 
damentally the  same  as  (Book  VI. 
Prop.  XX.) ;  for,  O  being  the  centre 
of  the  sphere,  a  sohd  angle  may  be 
conceived  as  formed  at  O  by  the  plane 
angles  AOB,  BOC,  COD,&c.,  and  the 
sum  of  these  angles  must  be  less  than 
four  right  angles  ;  which  is  exactly 
the  proposition  here  proved.     The  A 

demonstration  here  given  is  different  from  that  of  Book  VI. 
Prop.  XX. ;  both,  how^ever,  suppose  that  the  polygon  ABCDE 
is  convex,  or  that  no  side  produced  will  cut  the  figure. 


PROPOSITION  V.    THEOREM. 


b 


The  poles  of  a  great  circle  of  a  sphere,  are  the  extremities  of  that 
diameter  of  the  sphere  which  is  perpendicular  to  the  circle  ; 
and  these  extremities  are  also  the  poles  of  all  small  circles 
parallel  to  it. 

Let  ED  be  perpendic- 
ular to  the  great  circle 
AMB ;  then  will  E  and 
D  be  its  poles ;  as  also 
the  poles  of  the  parallel 
small  circles  HPI,FNG. 

For,  DC  being  per- 
pendicular to  the  plane 
AMB,  is  perpendicular 
to  all  the  straight  lines 
CA,  CM,  CB,  Redrawn 
through  its  foot  in  this 
plane  ;  hence  all  the  arcs 
DA,  DM,  DB,  &c.  are 
quarters  of  the  circumfe- 
rence.    So  likewise  are 

all  the  arcs  EA,  EM,  EB,  &c. ;  hence  the  points  D  and  E  are 
each  equally  distant  from  all  the  points  of  the  circumference 
AMB ;  hence,  they  are  the  poles  of  that  circumference  (Def.  7.), 

Again,  the  radius  DC,  perpendicular  to  the  plane  AMB,  is 
perpendicular  to  its  parallel  FNG ;  hence,  it  passes  through  O 
the  centre  of  the  circle  FNG  (Book  VIII.  Prop.  VII.  Cor.  4.)  ; 
hence,  if  the  oblique  lines  DF,  DN,  DG,  be  drawn,  these  ob- 
lique lines  will  diverge  equally  from  the  perpendicular  DO, 
and  will  themselves  be  equal.     But,  the  chords  being  equal, 


190 


GEOMETRY. 


the  arcs  are  equal ;  hence  the  point  D  is  the  pole  of  the  small 
circle  FNG  ;  and  for  like  reasons,  the  point  E  is  the  other  pole. 

Cor.  1.  Every  arc  DM, 
drawn  from  a  point  in 
the  arc  of  a  great  circle 
AMB  to  its  pole,  is  a  quar- 
ter of  the  circumference, 
which  for  the  sake  of 
brevity,  is  usually  named 
a  quadrant  :  and  this 
quadrant  at  the  same 
time  makes  a  right  angle 
with  the  arc  AM.  For, 
the  line  DC  being  per- 
pendicular to  the  plane 
AMC,  every  plane  DME, 
passing  through  the  line 
DC  is  perpendicular  to 
the  plane  AMC  (Book  VI.  Prop.  XVI.) ;  hence,  the  angle  of 
these  planes,  or  the  angle  AMD,  is  a  right  angle. 

Cor.  2.  To  find  the  pole  of  a  given  arc  AM,  draw  the  indefi- 
nite arc  MD  perpendicular  to  AM  ;  take  MD  equal  to  a  quad- 
rant ;  the  point  D  will  be  one  of  the  poles  of  the  arc  AM :  or 
thus,  at  the  two  points  A  and  M,  draw  the  arcs  AD  and  MD 
perpendicular  to  AM ;  their  point  of  intersection  D  will  be  the 
pole  required. 

Cor.  3.  Conversely,  if  the  distance  of  the  point  D  from  each 
of  the  points  A  and  M  is  equal  to  a  quadrant,  the  point  D  will 
be  the  pole  of  the  arc  AM,  and  also  the  angles  DAM,  AMD, 
will  be  right  angles. 

For,  let  C  be  the  centre  of  the  sphere  ;  and  draw  the  radii 
CA,  CD,  CM.  Since  the  angles  ACD,  MCD,  are  right  angles, 
the  line  CD  is  perpendicular  to  the  two  straight  lines  CA,  CM ; 
hence  it  is  perperpendicular  to  their  plane  (Book  VI.  Prop. 
IV,) ;  hence  the  point  D  is  the  pole  of  the  arc  AM  ;  and  conse- 
quently the  angles  DAM,  AMD,  are  right  angles. 

Scholium.  The  properties  of  these  poles  enable  us  to  describe 
arcs  of  a  circle  on  the  surface  of  a  sphere,  with  the  same 
facility  as  on  a  plane  surface.  It  is  evident,  for  instance,  that 
by  turning  the  arc  DF,  or  any  other  line  extending  to  the  same 
distance,  round  the  point  D,  the  extremity  F  will  describe  the 
small  circle  FNG ;  and  by  turning  the  quadrant  DFA  round 


'r  >'*■ 


BOOK  IX. 


191 


the  point  D,  its  extremity  A  will  describe. the  arc  of  the  great 
circle  AMB. 

If  the  arc  AM  were  required  to  be  produced,  and  nothing 
were  given  but  the  points  A  and  M  through  which  it  was  to 
pass,  we  should  first  have  to  determine  the  pole  D,  by  the 
intersection  of  two  arcs  described  from  the  points  A  and  M  as 
centres,  with  a  distance  equal  to  a  quadrant ;  the  pole  D  being 
found,  we  might  describe  the  arc  AM  and  its  prolongation, 
from  D  as  a  centre,  and  with  the  same  distance  as  before. 

In  fine,  if  it  be  required  from  a  given  point  P,  to  let  fall  a 
perpendicular  on  the  given  arc  AM ;  find  a  point  on  the  arc 
AM  at  a  quadrant's  distance  from  the  point  P,  which  is  done  by 
describing  an  arc  with  the  point  P  as  a  pole,  intersecting  AM  in  S : 
S  will  be  the  point  required,  and  is  the  pole  with  which  a  per- 
pendicular to  AM  may  be  described  passing  through  the  point  P. 


PROPOSITION  VI.    THEOREM. 


TTie  angle  formed  by  two  arcs  of  great  circles^  is  equal  to  the  an- 
gle formed  by  the  tangents  of  these  arcs  at  their  point  of  inter- 
section,  and  is  measured  by  the  arc  described  from  this  point 
of  intersection,  as  a  pole,  and  limited  by  the  sides,  produced  if 
necessary. 

Let  the  angle  BAG  be  formed  by  the  two   A^ 
arcs  AB,  AC  ;   then  will  it  be  equal  to  the      "" 
angle  FAG  formed  by  the  tangents  AF,  AG, 
and  be  measured  by  the  arc  DE,  described 
about  A  as  a  pole. 

For  the  tangent  AF,  drawn  in  the  plane 
of  the  arc  AB,  is  perpendicular  to  the  radius  ^ 
AO ;  and  the  tangent  AG,  drawn  in  the  plane 
of  the  arc  AC,  is  perpendicular  to  the  same 
radius  AO.  Hence  the  angle  FAG  is  equal 
to  the  angle  contained  by  the  planes  ABO, 
OAC  (Book  VI.  Def.  4.)  ;  which  is  that  of  HL 
the  arcs  AB,  AC,  and  is  called  the  angle  BAC. 

In  like  manner,  if  the  arcs  AD  and  AE  are  both  quadrants, 
the  lines  OD,  OE,  will  be  perpendicular  to  OA,  and  the  angle 
DOE  will  still  be  equal  to  the  angle  of  the  planes  AOD,  AOE  : 
hence  the  arc  DE  is  the  measure  of  the  angle  contained  by 
these  planes,  or  of  the  angle  CAB. 

Cor.  The  angles  of  spherical  triangles  may  be  compared 
together,  by  means  of  the  arcs  of  great  circles  described  from 
their  vertices  as  poles  and  included  between  their  sides :  hence 
it  is  easy  to  make  an  angle  of  this  kind  equal  to  a  given  angle. 


im 


GEOMETRY. 


Scholium.  Vertical  angles,  such 
as  ACO  and  BCN  are  equal ;  for 
either  of  them  is  still  the  angle 
formed  by  the  two  planes  ACB, 
OCN. 

It  is  farther  evident,  that,  in  the 
intersection  of  two  arcs  ACB,  OCN, 
the  two  adjacent  angles  ACO,  OCB, 
taken  together,  are  equal  to  two 
right  angles. 


PROPOSITION  VII.     THEOREM. 


If  from  the  vertices  of  the  three  angles  of  a  spherical  triangle,  as 
poles,  three  arcs  he  described  forming  a  second  triangle,  the 
vertices  of  the  angles  of  this  second  triangle,  will  he  respectively 
poles  of  the  sides  of  the  first. 

From  the  vertices  A,  B,  C, 
as  poles,  let  the  arcs  EF,  FD, 
ED,  be  described,  forming  on 
the  surface  of  the  sphere,  the 
triangle  DFE ;  then  will  the 
points  D,  E,  and  F,  be  respec- 
tively poles  of  the  sides  BC, 
AC,  AB. 

For,  the  point  A  being  the 
pole  of  the  arc  EF,  the  dis- 
tance AE  is  a  quadrant ;  the 
point  C  being  the  pole  of  the  arc  DE,  the  distance  CE  is  like- 
wise a  quadrant :  hencie  the  point  E  is  removed  the  length  of  a 
quadrant  from  each  of  the  points  A  and  C  ;  hence,  it  is  the 
pole  of  the  arc  AC  (Prop.  V.  Cor.  3.).  It  might  be  shown,  by 
the  same  method,  that  D  is  the  pole  of  the  arc  BC,  and  F  that 
of  the  arc  AB. 

Cor.  Hence  the  triangle  ABC  may  be  described  by  means 
of  DEF,  as  DEF  is  described  by  means  of  ABC.  Triangles 
so  described  are  called  polar  tmangles,  or  supplemental  tri- 
angles. 


BOOK  IX. 


193 


PROPOSITION  VIII.    THEOREM. 


The  same  supposition  continuing  as  in  the  last  Pi'oposition,  each 
angle  in  one  of  the  triangles,  will  he  measured  by  a  semicir- 
cumference,  minus  the  side  lying  opposite  to  it  in  the  other 
triangle. 

For,  produce  the  sides  AB, 
AC,  if  necessary,  till  they  meet 
EF,  in  G  and  H.  The  point  A 
being  the  pole  of  the  arc  GH, 
the  angle  A  will  be  measured 
by  that  arc  (Prop.  VL).  But 
the  arc  EH  is  a  quadrant,  and 
likewise  GF,  E  being  the  pole 
of  AH,  and  F  of  AG  ;  hence 
EH  -f  GF  is  equal  to  a  semi- 
circumference.  Now,  EH  + 
GF  is  the  same  as  EF+  GH  ;  hence  the  arc  GH,  which  mea- 
sures the  angle  A,  is  equal  to  a  semicircumference  minus  the 
side  EF.  In  like  manner,  the  angle  B  will  be  measured  by 
^circ, — DF  :  the  angle  C,  by  |  circ. — DE. 

And  this  property  must  be  reciprocal  in  the  two  triangles, 
since  each  of  them  is  described  in  a  similar  manner  by  means 
of  the  other.  Thus  we  shall  find  the  angles  D,  E,  F,  of  the  triangle  • 
DEFtobe  measured  respectively  by  icirc. — BC,  |  circ. — AC, 
I  circ— AB.  Thus  the  angle  D,  for  example,  is  measured  by 
the  arc  MI ;  but  MI  +  BC  =  MC  +  BI=i  circ. ;  hence  the  arc 
MI,  the  measure  of  D,  is  equal  to  \  circ— BC  ;  and  so  of  all 
the  rest. 

Scholium.  It  must  further  be  observed, 
that  besides  the  triangle  ]l)EF,  three  others 
might  be  formed  by  the  intersection  of 
the  three  arcs  DE,  EF,  DF.  But  the 
proposition  immediately  before  us  is  ap- 
plicable only  to  the  central  triangle, 
which  is  distinguished  from  the  other 
three  by  the  circumstance  (see  the  last 
figure)  that  the  two  angles  A  and  D  lie 
on  the  same  side  of  BC,  the  two  B  and  E  on  the  same  side  of 
AC,  and  the  two  C  and  F  on  the  same  side  of  AB. 

R  25 


194  GEOMETRY. 


PROPOSITION  IX.    THEOREM 

If  around  the  vertices  of  the  two  angles  of  a  given  spherical  tri- 
angle, as  poles,  the  circumferences  of  two'  circles  be  described 
which  shall  pass  through  the  third  angle  of  the  triangle;  if  then, 
through  the  other  point  in  which  these  circumferences  intersect 
and  the  two  first  angles  of  the  triangle,  the  arcs  of  great  cir- 
cles be  drawn,  the  triangle  thus  formed  will  have  all  its  parts 
equal  to  those  of  the  given  triangle. 

Let  ABC  be  the  given  triangle,  CED, 
DFC,  the  arcs  described  about  A  and  B 
as  poles ;  then  will  the  triangle  ADB  have 
all  its  parts  equal  to  those  of  ABC. 

For,  by  construction,  the  side  ADrr: 
AC,  DB=BC,  and  AB  is  common  ;  hence 
these  tvv^o  triangles  have  their  sides  equal, 
each  to  each.  We  are  now  to  show,  that 
the  angles  opposite  these  equal  sides  are 
also  equal. 

If  the  centre  of  the  sphere  is  supposed  to  be  at  O,  a  solid 
angle  may  be  conceived  as  formed  at  O  by  the  three  plane 
angles  AOB,  AOC,  BOC  ;  likewise  another  solid  angle  may  be 
conceived  as  formed  by  the  three  plane  angles  AOB,  AOD, 
BOD.  And  because  the  sides  of  the  triangle  ABC  are  equal 
to  those  of  the  triangle  ADB,  the  plane  angles  forming  the  one 
of  these  solid  angles,  must  be  equal  to  the  plane  angles  forming 
the  other,  each  to  each.  But  in  that  case  we  have  shown  that 
the  planes,  in  which  the  equal  angles  lie,  are  equally  inclined 
to  each  other  (Book  VI.  Prop.  XXI.) ;  hence  all  the  angles  of 
the  spherical  triangle  DAB  are  respectively  equal  to  those  of 
the  triangle  CAB,  namely,  DAB=BAC,  DBA=ABC,  and 
ADB=ACB;  hence  the  sides  and  the  angles  of  the  triangle 
ADB  are  equal  to  the  sides  and  the  angles  of  the  triangle  AC3. 

Scholium.  The  equality  of  these  triangles  is  not,  however, 
an  absolute  equality,  or  one  of  superposition ;  foi*  it  would  be 
impossible  to  apply  them  to  each  other  exactly,  unless  they 
•were  isosceles.  The  equality  meant  here  is  what  we  have 
already  named  an  equality  by  symmetry  ;  therefore  we  shall 
call  the  triangles  ACB,  ADB,  symmetrical  triangles. 


BOOK  IX.  195 


PROPOSITION  X.    THEOREM. 

Tivo  triangles  on  the  same  sphere^  or  on  equal  spheres^  are  equal 
in  all  their  parts,  when  two  sides  and  the  included  angle  of  the 
one  are  equal  to  two  sides  and  the  included  angle  of  the  othery 
each  to  each. 

Suppose  the  side  AB=EF,  the  side 
AC =EG,  and  the  angle  BAC^FEG ; 
then  will  the  two  triangles  be  equal 
in  all  their  parts.    " 

For,  the  triangle  EFG  may  be 
placed  on  the  triangle  ABC,  or  on 
ABD  symmetrical  with  ABC,  just  as 
two  rectilineal  triangles  are  placed  V^ 
upon  each  other,  when  they  have  an  B 
equal  angle  included  between  equal  sides.  Hence  all  the  parts 
of  the  triangle  EFG  will  be  equal  to  all  the  parts  of  the  trian- 
gle ABC  ;  that  is,  besides  the  three  parts  equal  by  hypothesis, 
we  shall  have  the  side  BC=FG,  the  angle  ABC=EFG,  and 
the  angle  ACB=EGF. 


PROPOSITION  XI.    THEOREM. 

Two  triangles  on  the  same  sphere,  or  on  equal  spheres,  are  equal 
in  all  their  parts,  when  two  angles  and  the  included  side  of  the 
one  are  equal  to  two  angles  and  the  included  side  of  the  other, 
each  to  each. 

For,  one  of  these  triangles,  or  the  triangle  symmetrical  with 
it,  may  be  placed  on  the  other,  as  is  done  in  the  corres- 
ponding case  of  rectilineal  triangles  (Book  I.  Prop.  VI.). 


PROPOSITION  XII.     THEOREM. 

If  two  triangles  on  the  same  sphere,  or  on  equal  spheres,  have  all 
their  sides  equal,  each  to  each,  their  angles  will  likewise  be 
equal,  each  to  each,  the  equal  angles  lying  opposite  the  equal 
sides. 


196 


GEOMETRY. 


This  truth  is  evident  from  Prop.  IX, 
where  it  was  shown,  that  with  three  given 
sides  AB,  AC,  BC,  there  can  only  be  two 
triangles  ACB,  ABD,  differing  as  to  the 
position  of  their  parts,  and  equal  as  to  the 
magnitude  of  those  parts.  Hience  tho§e 
two  triangles,  having  all  their  sides  re- 
spectively equal  in  both,  must  either  be 
absolutely  equal,  or  at  least  symmetrically 
so  ;  in  either  of  which  cases,  their  corres- 
ponding angles  must  be  equal,  and  lie  opposite  to  equal  sides. 


PROPOSITION  XIII.     THEOREM. 


In  every  isosceles  spherical  triangle,  the  angles  opposite  the  equal 
sides  are  equal ;  and  conversely,  if  two  angles  of  a  spherical 
tnangle  are  equal,  the  triangle  is  isosceles. 

First.  Suppose  the  side  AB=AC;  we  shall 
have  the  angle  C=B.  For,  if  the  arc  AD  be" 
drawn  from  the  vertex  A  to  the  middle  point 
D  of  the  base,  the  two  triangles  ABD,  ACD, 
will  have  all  the  sides  of  the  one  respectively 
equal  to  the  corresponding  sides  of  the  other,- 
namely,  AD  common,  BD  =  DC,  and  AB== 
AC  :  hence  by  the  last  Proposition,  their  an- 
gles will  be  equal;  therefore,  B  =  C. 

Secondly.  Suppose  the  angle  B  =  C  ;  we  shall  have  the  side 
AC=AB.  For,  if  not,  let  AB  be  the  greater  of  the  two  ;  take 
BO=iAC,  and  draw  OC.  The  two  sides  BO,  BC,  are  equal  to 
the  two  AC,  BC ;  the  angle  OBC,  contained  by  the  first  two 
is  equal  to  ACB  contained  by  the  second  two.  Hence  the 
two  triangles  BOC,  ACB,  have  all  their  other  parts  equal 
(Prop.  X.) ;  hence  the  angle  OCBi=ABC  :  but  by  hypothesis, 
the  angle  ABC  =r  ACB  ;  hence  we  have  OCB=ACB,  which  is 
absurd  ;  hence  it  is  absurd  to  suppose  AB  different  from  AC ; 
hence  the  sides  AB,  AC,  opposite  to  the  equal  angles  B  and  C, 
are  equal. 

Scholium.  The  same  demonstration  proves  the  anglie  BAD  = 
DAC,  and  the  angle  BDAirzADC.  Hence  the  two  last  are 
right  angles  ;  hence  the  are  drawn  from  the  vertex  of  an  isosceles 
spherical  triangle  to  the  middle  of  the  base,  is  at  right  angles  to 
that  base,  and  bisects  the  vertical  angle. 


BOOK  IX.  197 


PROPOSITION  XIV.    THEOREM. 


In  any  spherical  trianghy  the  greater  side  is  opposite  the  greater 
angle  ;  and  conversely ^  the  greater  angle  is  opposite  the  greater 
side. 

Let  the  angle  A  be  greater 
than  the  angle  B,  then  will  BC 
be  greater  than  AC  ;  and  con- 
versely, if  BC  is  greater  than 
AC,  then  will  the  angle  A  be  -^^ 
greater  than  B. 

First.  Suppose  the  angle  A>B  ;  make  the  angle  BAD=B  ; 
then  we  shall  have  AD=DB  (Prop.  XIII.) :  but  AD  +  DC  is 
greater  than  AC  ;  hence,  putting  DB  in  place  of  AD,  we  shall 
have  DB  +  DC,  or  BC>AC. 

Secondly.  If  we  suppose  BC  > AC,  the  angle  BAC  will  be 
greater  than  ABC.  For,  if  BAC  were  equal  to  ABC,  we 
should  have  BC=AC  ;  if -BAC  were  less  than  ABC,  we  should 
then,  as  has  justi)een  shown,  find  BC<AC.  Both  these  con- 
clusions are  false :  hence  the  angle  BAC  is  greater  than  ABC. 


PROPOSITION  XV.    THEOREM. 

If  two  triangles  on  the  same  sphere,  or  on  equal  spheres,  are 
mutually  equiangular,  they  will  also  he  mutually  equilateral. 

Let  A  and  B  be  the  two  given  triangles  ;  P  and  Q  their  polar 
triangles.  Since  the  angles  are  equal  in  the  triangles  A  and 
B,  the  sides  will  be  equal  in  their  polar  triangles  P  and  Q 
(Prop.  VIII.) :  but  since  the  triangles  P  and  Q  are  mutually 
evuilateral,  they  must  also  be  mutually  equiangular  (Prop. 
XII.)  ;  and  lastly,  the  angles  being  equal  in  the  triangles  P 
and  Q,  it  follows  that  the  sides  are  equal  in  their  polar  trian- 
gles A  and  B.  Hence  the  mutually  equiangular  triangles  A 
and  B  are  at  the  same  time  mutually  equilateral. 

Scholium.  This  proposition  is  not  applicable  to  rectilineal 
triangles ;  in  which  equality  among  the  angles  indicates  only 
proportionality  among  the  sides.  Nor  is  it  difficult  to  account 
for  the  difference  observable,  in  this  respect,  between  spherical 
and  rectilineal  triangles.     In  the  Proposition  now  before  us, 


198  GEOMETRY. 

as  well  as  in  the  preceding  ones,  which  treat  of  the  comparison 
of  triangles,  it  is  expressly  required  that  the  arcs  be  traced  on 
the  same  sphere,  or  on  equal  spheres.  Now  similar  arcs  are 
to  each  other  as  their  radii ;  hence,  on  equal  spheres,  two  tri- 
angles cannot  be  similar  without  being  equal.  Therefore  it  is 
not  strange  that  equality  among  the  angles  should  produce 
equality  among  the  sides. 

The  case  would  be  different,  if  the  triangles  were  drawn 
upon  unequal  spheres  ;  there,  the  angles  being  equal,  the  trian- 
gles would  be  similar,  and  the  homologous  sides  would  be  to 
each  other  as  the  radii  of  their  spheres. 


PROPOSITION  XVI.     THEOREM. 

The  sum  of  all  the  angles  in  any  spherical  triangle  i>  less  than 
six  right  angles,  and  greater  than  tioo. 

For,  in  the  first  place,  every  angle  of  a  spherical  triangle  is 
less  than  two  right  angles  :  hence  the  sum  of  all  the  three  is 
less  than  six  right  angles. 

Secondly,  the  measure  of  each  angle  of  a  spherical  triangle 
is  equal  to  the  semicircumference  minus  the  corresponding  side 
of  the  polar  triangle  (Prop.  VIII.) ;  hence  the  sum  of  all  the  three, 
is  measured  by  the  three  semicircumferences  minusih^  sum  of  all 
the  sides  of  the  polar  triangle.  Now  this  latter  sum  is  less  than  a 
circumference  (Prop,  III.) ;  therefore,  taking  it  away  from  three 
semicircumferences,  the  remainder  will  be  greater  than  one 
semicircumference,  which  is  the  measure  of  two  right  angles ; 
hence,  in  the  second  place,  the  sum  of  all  the  angles  of  a  sphe- 
rical triangle  is  greater  than  two  right  angles. 

Cor.  1 .  The  sum  of  all  the  angles  of  a  spherical  triangle  is 
not  constant,  like  that  of  all  the  angles  of  a  rectilineal  triangle  ; 
it  varies  between  two  right  angles  and  six,  without  ever  arriving 
at  either  of  these  limits.  Two  given  angles  therefore  do  not 
serve  to  determine  the  third.* 

Cor.  2.  A  spherical  triangle  may  have  two,  or  even  three  of 
its  angles  right  angles ;  also  two,  or  even  three  of  its  angles 
obtuse. 


BOOK  IX.  199 

Cor,  3.  If  the  triangle  ABC  is  hi-rectangular, 
in  other  words,  has  two  right  angles  B  and  C, 
the  vertex  A  will  be  the  pole  of  the  base  BC  ; 
and  the  sides  AB,  AC,  will  be  quadrants 
(Prop.  V.  Cor.  3.). 

If  the  angle  A  is  also  a  right  angle,  the  tri- 
angle ABC  will  be  tri-rectangular ;  its'  angles 
will  all  be  right  angles,  and  its  sides  quadrants, 
tri-rectangular  triangles  make  half  a  hemisphere,  four  make  a 
hemisphere,  and  the  tri-rectangular  triangle  is  obviously  con- 
tained eight  times  in  the  surface  of  a  sphere. 

Scholium.  In  all  the  preceding 
observations,  we  have  supposed,  in 
conformity  with  (Def.  1.)  that  sphe- 
rical triangles  have  always  each  of 
their  sides  less  than  a  semicircum- 
ference  ;  from  which  it  follows  that 
any  one  of  their  angles  is  always 
less  than  two  right  angles.  For,  if 
the  side  AB  is  less  than  a  semicir- 
cumference,  and  AC  is  so  likewise, 
both  those  arcs  will  require  to  be  E 

produced,  before  they  can  meet  in  D.  Now  the  two  angles 
ABC,  CBD,  taken  together,  are  equal  to  two  right  angles  ; 
hence  the  angle  ABC  itself,  is  less  than  two  right  angles. 

We  may  observe,  however,  that  some  spherical  triangles  do 
existr  in  which  certain  of  the  sides  are  greater  than  a  semicir- 
cumference,  and  certain  of  the  angles  greater  than  two  right 
angles.  Thus,  if  the  side  AC  is  produced  so  as  to  form  a  whole 
circumference  ACE,  the  part  which  remains,  after  subtracting 
the  triangle  ABC  from  the  hemisphere,  is  a  new  triangle  also 
designated  by  ABC,  and  having  AB,  BC,  AEDC  for  its  sides. 
Here,  it  is  plain,  the  side  AEDC  is  greater  than  the  semicir- 
cumferencc  AED  ;  and  at  the  same  time,  the  angle  B  opposite 
to  it  exceeds  two  right  angles,  by  the  quantity  CBD. 

The  triangles  whose  sides  and  angles  are  so  large,  have  been 
excluded  by  the  Definition  ;  but  the  only  reason  was,  that  the 
solution  of  them,  or  the  determmation  of  their  parts,  is  always 
reducible  to  the  solution  of  such  triangles  as  are  comprehended 
by  the  Definition.  Indeed,  it  is  evident  enough,  that  if  the  sides 
and  angles  of  the  triangle  ABC  are  known,  it  will  be  easy  to 
discover  the  angles  and  sides  of  the  triangle  which  bears  the 
same  name,  and  is  the  difference  between  a  hemisphere  and  the 
former  triangle. 


200  GEOMETRY. 


PROPOSITION  XVII.    THEOREM. 

The  surface  of  a  lune  is  to  the  surface  of  the  sphere,  as  the  angle 
of  this  lune,  is  to  four  right  angles,  or  as  the  arc  which  mea- 
sures that  angle,  is  to  the  circumference. 

Let  AMBN  be  a  lune  ;  then  will  its 
surface  be  to  the  surface  of  the  sphere 
as  the  angle  NCM  to  four  right  angles, 
or  as  the  arc  NM  to  the  circumference 
of  a  great  circle. 

Suppose,  in  the  first  place,  the  arc 
MN  to  be  to  the  circumference  MNPQ 
as  some  one  rational  number  is  to  ano- 
ther, as  5  to  48,  for  example.  The  cir- 
cumference MNPQ  being  divided  into 
48  equal  parts,  MN  will  contain  5  of  them  ;  and  if  the  pole  A 
were  joined  with  the  several  points  of  division,  by  as  many 
quadrants,  we  should  in  the  hemisphere  AMNPQ  have  48  tri- 
angles, all  equal,  because  all  their  parts  are  equal.  Hence  the 
whole  sphere  must  contain  96  of  those  partial  triangles,  the  lune 
AMBNA  will  contain  10  of  them ;  hence  the  lune  is  to  the 
sphere  as  10  is  to  96,  or  as  5  to  48,  in  other  words,  as  the  arc 
MN  is  to  the  circumference. 

If  the  arc  MN  is  not  commensurable  with  the  circumference, 
we  may  still  show,  by  a  mode  of  reasoning  frequently  exem- 
plified already,  that  in  that  case  also,  the  lune  is  to  the  sphere 
as  MN  is  to  the  circumference. 

Cor.  1.  Two  lunes  are  to  each  other  as  their  respective 
angles. 

Cor.  2.  It  was  shown  above,  that  the  whole  surface  of  the 
sphere  is  equal  to  eight  tri-rectangular  triangles  (Prop.  XVI. 
Cor.  3.)  ;  hence,  if  the  area  of  one  such  triangle  is  taken  for 
unity,  the  surface  of  the  sphere  will  be  represented  by  8.  This 
granted,  if  the  right  angle  be  assumed  equal  to  1,  the  surface 
of  the  lune  whose  angle  is  A  will  be  expressed  by  2A  ;  for, 

4  :  8  :  :  A  :  2A. 
Thus  we  have  here  two  different  unities  ;  one  for  angles,  being 
the  right  angle  ;  the  other  for  surfaces,  being  the  tri-rectangu- 
lar spherical  triangle,  or  the  triangle  whose  angles  are  all  right 
angles,  and  whose  sides  are  quadrants. 

Scholium.  The  spherical  ungula,  bounded  by  the  planes 
AMB,  ANB,  is  to  the  whole  solid  sphere,  as  the  angle.  A  is  to 


BOOK  IX.  201 

four  right  angles.  For,  the  lunes  being  equal,  the  spherical 
ungulas  will  also  be  equal ;  hence  two  spherical  ungulas  are  to 
each  other,  as  the  angles  formed  by  the  planes  which  bound 
them. 


PROPOSITION  XVIII.     THEOREM. 
T\oo  symmetrical  spherical  triangles  are  equivalent. 

Let  ABC,  DEP,  be  two  symmetri- 
cal triangles,  that  is  to  say,  two  tri- 
angles having  their  sides  AB=DE, 
AC=DF,  CB=EF,  and  yet  incapa- 
ble of  coinciding  with  each  other  ;  /  I  q  p  / 
we  are  to  show  that  the  surface  ABC  *'         ^"* 

is  equal  to  the  surface  DEF. 

Let  P  be  the  pole  of  the  small 
circle  passing  through  the  three  points 
A,  B,  C  ;*  from  this  point  draw  the 
equal  arcs  PA,  PB,  PC  (Prop.  V.)  ;  at  the  point  F,  make  the 
angle  DFQrzACP,  the  arc  FQ=r:CP ;  and  draw  DQ,  EQ. 

The  sides  DF,  FQ,  are  equal  to  the  sides  AC,  CP  ;  the  an- 
gle DFQ= ACP :  hence  the  two  triangles  DFQ,  ACP  are  equal 
in  all  their  parts  (Prop.  X.)  ;  hence  the  side  DQ=:AP,  and  the 
angle  DQF=APC, 

In  the  proposed  triangles  DFE,  ABC,  the  angles  DFE,  ACB, 
opposite  to  the  equal  sides  DE,  AB,  being  equal  (Prop.  XII.). 
if  the  angles  DFQ,  ACP,  which  are  equal  by  construction,  be 
taken  away  from  them,  there  will  remain  the  angle  QFE,  equal 
to  PCB.  Also  the  sides  QF,  FE,  are  equal  to  the  sides  PC, 
CB ;  hence  the  two  triangles  FQE,  CPB,  are  equal  in  all  their 
parts ;  hence  the  side  QE^PB,  and  the  angle  FQE  =  CPB. 

Now,  the  triangles  DFQ,  ACP,  which  have  their  sides  re- 
spectively equal,  are  at  the  same  time  isosceles,  and  capable  of 
<5oinciding,  when  applied  to  each  other ;  for  having  placed  AB 
on  its  equal  DF,  the  equal  sides  will  fall  on  each  other,  and 
thus  the  two  triangles  will  exactly  coincide  :  hence  they  are 
equal ;  and  the  surface  DQF=:APC.  '  For  a  like  reason,  the 
surface  FQE  =  CPB,  and  the  surface  DQE=:APB ;  hence  we 


*  The  circle  which  passes  through  the  three  points  A,  B,  C,  or  which  cir- 
cumscribes  the  triangle  ABC,  can  only  be  a  small  circle  of  the  sphere ;  for  if 
it  were  a  great  circle,  the  three  sides  AB,  BC,  AC,  would  lie  in  one  plane,  and 
the  triangle  ABC  would  be  reduced  to  one  of  its  sides. 

26 


202 


GEOMETRY. 


have    DQF+FQE— DQE=APC  +  CPB— APB,  or   DFE  = 

ABC ;  hence  the  two  symmetrical  triangles  ABC,  DEF  are 
equal  in  surface. 

Scholium.  The  poles  P  and  Q 
might  lie  within  triangles  ABC, 
DEF:  in  which  case  it  would  be 
requisite  to  add  the  three  triangles 
DQF,  FQE,  DQE,  together,  in  or- 
der to  make  up  the  triangle  DEF ; 
and  in  like  manner,  to  add  the  three 
triangles  APC,  CPB,  APB,  together, 
in  order  to  make  up  the  triangle 
ABC  :  in  all  other  respects,  the  de- 
monstration and  the  result  would  still  be  the  same 


PROPOSITION  XIX.     THEOREM. 


If  the  circumferences  of  two  great  circles  intersect  each  other  on 
the  surface  of  a  hemisphere^  the  sum  of  the  opposite  triangles 
thus  formed,  is  equivalent  to  the  surface  of  a  lune  whose  angle 
■  is  equal  to  the  angle  formed  by  the  circles. 

Let  the  circumferences  AOB,  COD,  in- 
tersect on  the  hemisphere  OACBD  ;  then 
will  the  opposite  triangles  AOC,  BOD  be 
equal  to  the  lune  whose  angle  is  BOD. 

For,  producing  the  arcs  OB,  OD,  on 
the  other  hemisphere,  till  they  meet  in 
N,  the  arc  OBN  will  be  a  semicircumfe- 
rence,  and  AOB  one  also ;  and  taking  OB 
from  both,  w^e  shall  h'ave  BN=AO.  For  a  like  reason,  we  have 
DN=CO,  and  BD= AC.  Hence  the  two  triangles  AOC,  BDN, 
have  their  three  sides  respectively  equal ;  besides  they  are  so 
placed  as  to  be  symmetrical ;  hence  they  are  equal  in  surface 
(Prop.  XVIII.) ,  but  the  sum  of  the  triangles  BDN,  BOD,  is 
equivalent  to  the  lune  OBNDO  whose  angle  is  BOD :  hence 
AOC  4-  BOD  is  equivalent  to  the  lune  whose  angle  is  BOD. 

Scholium.  It  is  likewise  evident  that  the  two  spherical  pyra- 
mids, which  have  the  triangles  AOC,  BOD,  for  bases,  are 
together  equivalent  to  the  spherical  ungula  whose  angle  is 
BOD. 


BOOK  IX. 


203 


PROPOSITION  XX.    THEOREM. 

The  surface  of  a  spherical  triangle  is  measured  hy  the  excess  of 
the  sum  of  its  three  angles  above  two  right  angles. 

Let  ABC  be  the  proposed  triangle  :  pro- 
duce its  sides  till  they  meet  the  great  circle 
DEFG  drawn  at  pleasure  without  the  tri- 
angle. By  the  last  Theorem,  the  two  tri- 
angles ADE,  AGH,  are  together  equiva- 
lent to  the  lune  whose  angle  is  A,  and  which 
is  measured  by  2A  (Prop.  XVII.  Cor.  2.). 
Hence  we  have  ADE  +  AGH=2A  ;  and 
for  a  like  reason,  BGF  +  BID=2B,  and 
CIH  +  CFE=2C.  But  the  sum  of  those  six  triangles  exceeds 
the  hemisphere  by  twice  the  triangle  ABC,  and  the  hemisphere 
is  represented  by  4  ;  therefore  twice  the  triangle  ABC  is  equal 
to  2A  +  2B  +  2C — 4  ;  and  consequently,  once  ABC=A  +  B  + 
C — 2  ;  hence  every  spherical  triangle  is  measured  by  the  sum 
of  all  its  angles  minus  two  right  angles. 

Cor.  1.  However  many  right  angles  there  be  contained  in 
this  measure,  just  so  many  tri-rectangular  triangles,  or  eighths 
of  the  sphere,  will  the  proposed  triangle  contain  (Prop.  XVII. 
Cor.  2.).  If  the  angles,  for  example,  are  each  equal  to  f  of  a 
right  angle,  the  three  angles  will  amount  to  4  right  angles,  and 
the  proposed  triangle  will  be  represented  by  4—2  or  2  ;  there- 
fore it  will  be  equal  to  two  tri-rectangular  triangles,  or  to  the 
fourth  part  of  the  whole  surface  of  the  sphere. 

Cor.  2.  The  spherical  triangle  ABC  is  equivalent  to  the  lune 


whose  angle  is 


A  +  B  +  C 


•1  ;  likewise  the  spherical  pyramid, 


which  has  ABC  for  its  base,  is  equivalent  to  the  spherical 
ungula  whose  angle  is 1. 


Scholium.  While  the  spherical  triangle  ABC  is  compared 
with  the  tri-rectangular  triangle,  the  spherical  pyramid,  which 
has  ABC  for  its  base,  is  compared  with  the  tri-rectangular 
pyramid,  and  a  similar  proportion  is  found  to  subsist  between 
them.  The  solid  angle  at  the  vertex  of  the  pyramid,  is  in  like 
manner  compared  with  the  solid  angle  at  the  vertex  of  the  tri- 
rectangular  pyramid.  These  comparisons  are  founded  on  the 
coincidence  of  the  corresponding  parts.     If  the  bases  of  the 


204  GEOMETRY. 

pyramids  coincide,  the  pyramids  themselves  will  evidently  co- 
incide, and  likewise  the  solid  angles  at  their  vertices.  From 
this,  some  consequences  are  deduced. 

First.  Two  triangular  spherical  pyramids  are  to  each  other 
as  their  bases  :  and  since  a  polygonal  pyramid  may  always  be 
divided  into  a  certain  number  of  triangular  ones,  it  follows  that 
any  two  spherical  pyramids  are  to  each  other,  as  the  polygons 
which  form  their  bases. 

Second.  The  solid  angles  at  the  vertices  of  these  pyramids,  are 
also  as  their  bases  ;  hence,  for  comparing  any  two  solid  angles, 
we  have  merely  to  place  their  vertices  at  the  centres  of  two 
equal  spheres,  and  the  solid  angles  will  be  to  each  other  as  the 
spherical  polygons  intercepted  between  their  planes  or  faces. 

The  vertical  angle  of  the  tri-rectangular  pyramid  is  formed 
by  three  planes  at  right  angles  to  each  other  :  this  angle,  which 
may  be  called  a  right  solid  angle,  will  serve  as  a  very  natural 
unit  of  measure  for  all  other  solid  angles.  And  if  so,  the  same 
number,  that  exhibits  the  area  of  a  spherical  polygon,  will  exhi- 
bit the  measure  of  the  corresponding  solid  angle.  If  the  area 
of  the  polygon  is  f ,  for  example,  in  other  words,  if  the  polygon 
is  f  of  the  tri-rectangular  polygon,  then  the  corresponding  solid 
angle  will  also  be  |  of  the  right  solid  angle. 


PROPOSITION  XXI.    THEOREM. 

The  surface  of  a  spherical  polygon  is  measured  by  the  sum  of  all 
its  angles,  minus  two  right  angles  multiplied  by  the  numbet 
of  sides  in  the  polygon  less  two. 

From  one  of  the  vertices  A,  let  diago- 
nals AC,  AD  be  drawn  to  all  the  other 
vertices  ;  the  polygon  ABCDE  will  be 
divided  into  as  many  triangles  minus 
two  as  it  has  sides.  But  the  surface  of 
each  triangle  is  measured  by  the  sum  of 
all  its  angles  minus  two  right  angles  ; 
and  the  sum  of  the  angles  in  all  the  tri- 
angles is  evidently  the  same  as  that  of  all  tlie  angles  in  the  po- 
lygon ;  hence,  the  surface  of  the  polygon  is  equal  to  the  sum  of 
all  its  angles  diminished  by  twice  as  many  right  angles  as  it  has 
sides  minus  two. 

Scholium.    Let  s  be  the  sum  of  all  the  angles  in  a  spherical 
polygon,  n  the  number  of  its  sides ;  the  right  angle  being  taken 
for  unity,  the  surface  of  the  polygon  will  be  measured  by 
s— 2.(n— -2),  or  5— 2  71  +  4. 


APPENDIX. 


THE  REGULAR  POLYEDRONS. 

A  regular  polyedron  is  one  whose  faces  are  all  equal  regular 
polygons,  and  whose  solid  angles  are  all  equal  to  each  other. 
There  are  five  such  polyedrons. 

First.  If  the  faces  are  equilateral  triangles,  polyedrons  may 
be  formed  of  them,  having  solid  angles  contained  by  three  of 
those  triangles,  by  four,  or  by  five :  hence  arise  three  regular 
bodies,  the  tetraedron,  the  octaedron,  the  icosaedron.  No  other 
can  be  formed  with  equilateral  triangles  ;  for  six  angles  of  such 
a  triangle  are  equal  to  four  right  angles,  and  cannot  form  a 
solid  angle  (Book  VI.  Prop.  XX.). 

Secondly.  If  the  faces  are  squares,  their  angles  may  be  ar- 
ranged by  threes :  hence  results  the  hexaedron  or  cube.  Four 
angles  of  a  square  are  equal  to  four  right  angles,  and  cannot 
form  a  solid  angle. 

Thirdly.  In  fine,  if  the  faces  are  regular  pentagons,  their 
angles  likewise  may  be  arranged  by  threes :  the  regular  dode- 
caedron  will  result. 

We  can  proceed  no  farther  :  three  angles  of  a  regular  hexa- 
gon are  equal  to  four  right  angles ;  three  of  a  heptagon  are 
greater. 

Hence  there  can  only  be  five  regular  polyedrons ;  three  formed 
with  equilateral  triangles,  one  with  squares,  and  one  with  pen- 
tagons.     ^ 

Construction  of  the  Tetraedron. 

Let  ABC  be  the  equilateral  triangle 
which  is  to  form  one  face  of  the  tetrae- 
dron. At  the  point  O,  the  ceotre  of  this 
triangle,  erect  OS  perpendicular  to  the 
plane  ABC  ;  terminate  this  perpendicular 
in  S,  so  that  AS=AB;  draw  SB,  SC : 
the  pyramid  S-ABC  will  be  the  tetrae- 
dron required. 

For,  by  reason  of  the  equal  distances 
OA,  OB,  OC,  the  oblique  lines  SA,  SB,  SC,  are  equally  re- 


S 


206 


APPENDIX. 


m 


moved  from  the  perpendicular  SO,  and 
consequently  equal  (Book  VI.  Prop.  V.). 
One  of  them  SA=AB ;  hence  the  four 
faces  of  the  pyramid  S-ABC,  are  trian- 
gles, equal  to  the  given  triangle  ABC. 
And  the  solid  angles  of  this  pyramid 
are  all  equal,  because  each  of  them  is 
formed  by  three  equal  plane  angles: 
hence  this  pyramid  is  a  regular  tetrae- 
dron. 


Construction  of  the  Hexaedron 

Let  ABCD  be  a  given  square.  On  the 
base  ABOD,  construct  a  right  prism  whose 
altitude  AE  shall  be  equal  to  the  side  AB. 
The  faces  of  this  prism  will  evidently  be 
equal  squares  ;  and  its  solid  angles  all  equal, 
each  being  formed  with  three  right  angles  : 
hence  this  prism  is  a  regular  hexaedron  or 
cube. 


The  following  propositions  can  be  easily  proved. 

1.  Any  regular  polyedron  may  be  divided  into  as  many 
regular  pyramids  as  the  polyedron  has  faces ;  the  common 
vertex  of  these  pyramids  will  be  the  centre  of  the  polyedron  ; 
and  at  the  same  time,  that  of  the  inscribed  and  of  the  circum- 
scribed sphere. 

2.  The  solidity  of  a  regular  polyedron  is  equal  to  its  sur- 
face multiplied  by  a  third  part  of  the  radius  of  the  inscribed 
sphere. 

3.  Two  regular  polyedrons  of  the  same  name,  are  two  simi- 
lar solids,  and  their  homologous  dimensions  are  proportional ; 
hence  the  radii  of  the  inscribed  or  the  circumscribed  spheres 
are  to  each  other  as  the  sides  of  the  polyedrons. 

4.  If  a  regular  polyedron  is  inscribed  in  a  sphere,  the  planes 
drawn  from  the  centre,  through  the  different  edges,  will  divide 
the  surface  of  the  sphere  into  as  many  spherical  polygons,  all 
equal  and  similar,  as  the  polyedron  has  faces. 


PLANE  TRIGONOMETRY.        207 


PLANE  TRIGONOMETRY. 

In  every  triangle  there  are  six  parts :  three  sides  and  three 
angles.  These  parts  are  so  related  to  each  other,  that  if  a 
certain  number  of  them  be  known  or  given,  the  remaining 
ones  can  be  determined. 

Plane  Trigonometry  explains  the  methods  of  finding,  by  cal- 
culation, the  unknown  parts  of  a  rectilineal  triangle,  when 
a  sufficient  number  of  the  six  parts  are  given. 

When  three  of  the  six  parts  are  known,  and  one  of  them  is  a 
side,  the  remaining  parts  can  always  be  found.  If  the  three 
angles  were  given,  it  is  obvious  that  the  problem  would  be  in- 
determinate, since  all  similar  triangles  would  satisfy  the  con- 
ditions. 

It  has  already  been  shown,  in  the  problems  annexed  to  Book 
III.,  how  rectilineal  triangles  are  constructed  by  means  of  three 
given  parts.  But  these  constructions,  which  are  called  graphic 
methods^  though  perfectly  correct  in  theory,  would  give  only 
a  moderate  approximation  in  practice,  on  account  of  the  im- 
perfection of  the  instruments  required  in  constructing  them. 
Trigonometrical  methods,  on  the  contrary,  being  independent 
of  all  mechanical  operations,  give  solutions  with  the  utmost 
accuracy. 

They  are  founded  upon  the  properties  of  lines  called  trigo- 
nometrical lines,  which  furnish  a  very  simple  mode  of  express- 
ing the  relations  between  the  sides  and  angles  of  triangles. 

We  shall  first  explain  the  properties  of  those  lines,  and  the 
principal  formulas  derived  from  them  ;  formulas  which  are  of 
great  use  in  all  the  branches  of  mathematics,  and  which  even 
furnish  means  of  improvement  to  algebraical  analysis.  We 
shall  next  apply  those  results  to  the  solution  of  rectilineal  tri- 
angles. 


DIVISION  OF  TKfi  CIRCUMFERENCE. 

I.  For  the  purposes  of  trigonometrical  calculation,  the  cir- 
cumference of  the  circle  is  divided  into  360  equal  parts,  called 
degrees ;  each  degree  into  60  equal  parts,  called  minutes ;  and 
each  minute  into  60  equal  parts,  called  seconds. 

The  semicircumference,  or  the  measure  of  two  right  angles, 
contains  180  degrees  ;  the  quarter  of  the  circumference,  usually 
denominated  the  quadrant,  and  which  measures  the  right  an- 
gle, contains  90  degrees. 

XL  Degrees,  minutes,  and  seconds,  are  respectively  desig- 


208 


PLANE  TRIGONOMETRY. 


nated  by  the  characters  :  «, ',  " :  thus  the  expression  16°  6'  15" 
represents  an  arc,  or  an  angle,  of  16  degrees,  6  minutes,  and 
15  seconds. 

III.  The  complement  of  an  angle,  or  of  an  arc,  is  what  re- 
mains after  taking  that  angle  or  that  arc  from  90°.  Thus  the 
complement  of  25°  40'  is  equal  to  90° — 25°  40'=64°  20^ ;  and 
the  complement  of  12°  4'  32"  is  equal  to  90° — 12^  4'  32"  =  77° 
55'  28". 

In  general,  A  being  any  angle  or  any  arc,  90° — A  is  the  com- 
plement of  that  angle  or  arc.  If  any  arc  or  angle  be  added 
to  its  complement,  the  sum  will  be  90°.  Whence  it  is  evident 
that  if  the  angle  or  arc  is  greater  than  90°,  its  complement  will 
be  negative.  Thus,  the  complement  of  160°  34'  10"  is  — 70° 
34'  10".  In  this  case,  the  complement,  taken  positively,  would 
be  a  quantity,  which  being  subtracted  from  the  given  angle  or 
arc,  the  remainder  would  be  equal  to  90°. 

The  two  acute  angles  of  a  right-angled  triangle,  are  together 
equal  to  a  right  angle  ;  they  are,  therefore,  complements  of  each 
other. 

IV.  The  supplement  of  an  angle,  or  of  an  arc,  is  what  re- 
mains after  taking  that  angle  or  arc  from  180°.  Thus  A  being 
any  angle  or  arc,  180° — A  is  its  supplement. 

In  any  triangle,  either  angle  is  the  supplement  of  the  sum  of 
the  two  others,  since  the  three  together  make  180°. 

If  any  arc  or  angle  be  added  to  its  supplement,  the  sum  will 
be  180°.  Hence  if  an  arc  or  angle  be  greater  than  180°,  its 
supplement  will  be  negative.  Thus,  the  supplement  of  200°. 
is  — 20°.  The  supplement  of  any  angle  of  a  triangle,  or  indeed 
of  the  sum  of  either  two  angles,  is  always  positive. 


GENERAL  IDEAS  RELATING  TO  THE  TRIGONOMERICAL  LINES 


V.  The  sine  of  an  arc  is 
the  perpendicular  let  fall  from 
one  extremity  of  the  arc,  on 
the  diameter  which  passes 
through  the  other  extremity. 
Thus,  MP  is  the  sine  of  the 
arc  AM,  or  of  the  angle  ACM. 

The  tangent  of  an  arc  is  a 
line  touching  the  arc  at  one 
extremity,  and  limited  by  the 
prolongation  of  the  diameter 
which  passes  through  the 
other  extremity.  Thus  AT  is 
the  tangent  of  the  arc  AM, 
or  of  the  angle  ACM. 


S' 


N 

tC 

q"^ 

s 

p 

r 

/ 

r    \ 

/ 

A 

A 

s 

\ 

J^^' 

^ 

h^ 

V 

E 


PLANE  TRIGONOMETRY.  209 

The  secant  of  an  arc  is  the  line  drawn  from  the  centre  of 
the  circle  through  one  extremity  of  the  arc  and  limited  by  the 
tangent  drawn  through  the  other  extremity.  Thus  CT  is  the 
secant  of  the  arc  AM,  or  of  the  angle  ACM. 

The  versed  sine  of  an  arc,  is  the  part  of  the  diameter  inter- 
cepted between  one  extremity  of  the  arc  and  the  foot  of  the 
sine.  Thus,  AP  is  the  versed  sine  of  the  arc  AM,  or  the  smgle 
ACM. 

These  four  lines  MP,  AT,  CT,  AP,  are  dependent  upon  the 
arc  AM,  and  are  always  determined  by  it  and  the  radius  ;  they 
are  thus  designated : 

MP=sin  AM,  or  sin  ACM, 
ATirrtangAM,  or  tang  ACM, 
CTi^secAM,  or  sec  ACM, 
AP=ver-sin  AM,  or  ver-sin  ACM. 
VI.  Having  taken  the  arc  AD  equal  to  a  quadrant,  from  the 
points  M  and  D  draw  the  lines  MQ,  DS,  perpendicular  to  the 
radius  CD,  the  one  terminated  by  that  radius,  the  other  termi- 
nated by  the  radius  CM  produced ;  the  lines  MQ,  DS,  and  CS, 
will,  in  like  manner,  be  the  sine,  tangent,  and  secant  of  the  arc 
MD,  the  complement  of  AM.     For  the  sake  of  brevity,  they 
are  called  the  cosine^  cotangent,  and  cosecant,  of  the  arc  AM, 
and  are  thus  designated  : 

MQ=:cosAM,  or  cos  ACM, 
DSmcot  AM,  or  cot  ACM, 
CS=cosec  AM,  or  cosec  ACM. 
In  general,  A  being  any  arc  or  angle,  we  have 
cos  A=sin  (90° — A), 
cot  A = tang  (90° — A), 
cosec  A  =: sec  (90° — A). 
The  triangle  MQC  is,  by  construction,  equal  to  the  triangle 
CPM  ;  consequently  CPrrMQ  :  hence  in  the  right-angled  tri- 
angle CMP,  whose  hypothenuse  is  equal  to  the  radius,  the  two 
sides  MP,  CP  are  the  sine  and  cosine  of  the  arc  AM :  hence, 
the  cosine  of  an  arc  is  equal  to  that  part  of  the  radius  inter- 
cepted between  the  centre  and  foot  of  the  sine. 

The  triangles  CAT,  CDS,  are  similar  to  the  equal  triangles 
CPM,  CQM ;  hence  they  are  similar  to  each  other.  From 
these  principles,  we  shall  very  soon  deduce  the  different  rela- 
tions which  exist  between  the  lines  now  defined  :  before  doing 
so,  however,  we  must  examine  the  changes  which  those  lines 
undergo,  when  the  arc  to  which  they  relate  increases  from  zero 
to  180°. 

The  angle  ACD  is  called  the  first  quadrant ;  the  angle  DCB, 
the  second  quadrant ;  the  angle  BCE,  the  third  quadrant ;  and 
the  angle  ECA,  the  fourth  quadrant 

S*  27 


210 


PLANE  TRIGONOMETRY. 


VII.  Suppose  one  extrem- 
ity of  the  arc  remains  fixed  in 
A,  while  the  other  extremity, 
marked  M,  runs  successively 
throughout  the  whole  extent 
of  the  semicircumference, 
from  A  to  B  in  the  direction 
ADB. 

When  the  point  M  is  at  A, 
or  when  the  arc  AM  is  zero, 
the  three  points  T,  M,  P,  are 
confounded  with  the  point  A  ; 
whence  it  appears  that  the 
sine  and  tangent  of  an  arc 

zero,  are  zero,  and  the  cosine  and  secant  of  this  same  arc,  are 
each  equal  to  the  radius.  Hence  if  R  represents  the  radius  oi 
the  circle,  we  have 

sin  0=0,  tang  0=0,  cos  0=R,  secO=R. 

VIII.  As  the  point  M  advances  towards  D,  the  sine  increases, 
and  so  likewise  does  the  tangent  and  the  secant ;  but  the  cosine, 
the  cotangent,  and  the  cosecant,  diminish. 

When  the  point  M  is  at  the  middle  of  AD,  or  when  the  arc 
AM  is  45°,  in  which  case  it  is  equal  to  its  complement  MD, 
the  sine  MP  is  equal  to  the  cosine  MQ  or  CP  ;  and  the  trian- 
gle  CMP,  having  become  isosceles,  gives  the  proportion 
MP  :  CM  :  :  1  :   V2, 


or 


sm  45^ 


R 


Hence 


1  :  V2. 
R 


sin  45°= cos  45o=-;^=lR\/2. 


In  this  same  case,  the  triangle  CAT  becomes  isosceles  and 
equal  to  the  triangle  CDS  ;  whence  the  tangent  of  45°  and  its 
cotangent,  are  each  equal  to  the  radius,  and  consequently  we 
have 

tang  45°  =  cot  4!r=R. 

IX.  The  arc  AM  continuing  to  increase,  the  sine  increases 
till  M  arrives  at  D  ;  at  which  point  the  sine  is  equal  to  the  ra- 
dius, and  the  cosine  is  zero.     Hence  we  have 

sin  90° = R,    cos  90° = 0  ; 

and  it  may  be  observed,  that  these  values  are  a  consequence 

of  the  values  already  found  for  the  sine  and  cosine  of  the  arc 

zero  ;  because  the  complement  of  90^  being  zero,  we  have 

sin  90°=:cos  0°r=R,  and 

cos  90°=zsin  0°=0. 


PLANE  TRIGONOMETRY.  211 

As  to  the  tangent,  it  increases  very  rapidly  as  the  point  M 
approaches  D  ;  and  finally  when  this  point  reaches  D,  the  tan- 
gent properly  exists  no  longer,  because  the  lines  AT,  CD, 
being  parallel,  cannot  meet.  This  is  expressed  by  saying  that 
the  tangent  of  90°  is  infinite  ;  and  we  write  tang  90°=  ao 
The  complement  of  90''  being  zero,  we  have 

tang  O=cot  90O  and  cot  Orrtang  90o. 

Hence       cot  90° =0,  and  cot  0=ao  . 

X.  The  point  M  continuing  to  advance  from  D  towards  B, 
the  sines  diminish  and  the  cosines  increase.  Thus  M'P'  is  the 
sine  of  the  arc  AM',  and  M'Q,  or  CP'  its  cosine.  But  the  arc 
M'B  is  the  supplement  of  AM',  since  AM'  +  M'B  is  equal  to  a 
semicircumference  ;  besides,  if  M'M  is  drawn  parallel  to  AB, 
the  arcs  AM,  BM',  which  are  included  between  parallels,  will 
evidently  be  equal,  and  likewise  the  perpendiculars  or  sines 
MP,  M'P'.  Hence,  the  sine  of  an  arc  or  of  an  angle  is  equal  to 
the  sine  of  the  supplement  of  that  arc  or  angle. 

The  arc  or  angle  A  has  for  its  supplement  180° — ^A;  hence 
generally,  we  have 

sin  A = sin  (ISO^ — A.) 
The  same  property  might  also  be  expressed  by  the  equation 

sin  (90°  +  B)  =  sin  (90°-— B), 
B  being  the  arc  DM  or  its  equal  DM', 

XI.  The  same  arcs  AM,  AM',  which  are  supplements  of 
each  other,  and  which  have  equal  sines,  have  also  equal  co- 
sines CP,  CP' ;  but  it  must  be  observed,  that  these  cosines  lie 
in  different  directions.  The  line  CP  which  is  the  cosine  of  the 
arc  AM,  has  the  origin  of  its  value  at  the  centre  C,  and  is  esti- 
mated in  the  direction  from  C  towards  A ;  while  CP',  the  cosine 
of  AM'  has  also  the  origin  of  its  value  at  C,  but  is  estimated  in 
a  contrary  direction,  from  C  towards  B. 

Some  notation  must  obviously  be  adopted  to  distinguish  the 
one  of  such  equal  lines  from  the  other  ;  and  that  they  may  both 
be  expressed  analytically,  and  in  the  same  general  formula,  it  is 
necessary  to  consider  all  lines  which  are  estimated  in  one  di- 
rection as  positive,  and  those  which  are  estimated  in  the  con- 
trary direction  as  negative.  If,  therefore,  the  cosines  which 
are  estimated  from  C  towards  A  be  considered  as  positive, 
those  estimated  from  C  towards  B,  must  be  regarded  as  nega- 
tive.    Hence,  generally,  we  shall  have, 

cos  A=— cos  (180°--A) 
that  is,  the  cosine  of  an  arc  or  angle  is  equal  to  the  cosine  of  its 
supplement  taken  negatively. 

The  necessity  of  changing  the  algebraic  sign  to  correspond 


212 


PLANE  TRIGONOMETRY. 


with  the  change  of  direction 
in  the  trigonometrical  Hne, 
may  be  illustrated  by  the  fol- 
lowing example.  The  versed 
sine  AP  is  equal  to  the  radius 
CA  minus  CP  the  cosine  AM  : 
that  is, 
ver-sin  AM=R — cos  AM. 
Now  when  the  arc  AM  be- 
comes AM'  the  versed  sine 
AP,  becomes  A  F,  that  is  equal 
to  R  +  CF.  But  this  expression 
cannot  be  derived  from  the 
formula, 

ver-sin  AMrrR — cos  AM, 
unless  we  suppose  the  cosine  AM  to  become  negative  as  soon 
as  the  arc  AM  becomes  greater  than  a  quadrant. 

At  the  point  B  the  cosine  becomes  equal  to  — R ;  that  is, 
cos  180°=— R. 

For  all  arcs,  such  as  ADBN',  which  terminate  in  the  third 
quadrant,  the  cosine  is  estimated  from  C  towards  B,  and  is 
consequently  negative.  At  E  the  cosine  becomes  zero,  and  for 
all  arcs  which  terminate  in  the  fourth  quadrant  the  cosines  are 
estimated  from  C  towards  A,  and  are  consequently  positive. 

The  sines  of  all  the  arcs  which  terminate  in  the  first  and 
second  quadrants,  are  estimated  above  the  diameter  BA,  while 
the  sines  of  those  arcs  which  terminate  in  the  third  and  fourth 
quadrants  are  estimated  below  it.  Hence,  considering  the 
former  as  positive,  we  must  regard  the  latter  as  negative. 

XII.  Let  us  now  see  what  sign  is  to  be  given  to  the  tangent 
of  an  arc.  The  tangent  of  the  arc  AM  falls  above  the  line  BA, 
and  we  have  already  regarded  the  lines  estimated  in  the  direc- 
tion AT  as  positive  :  therefore  the  tangents  of  all  arcs  which 
terminate  in  the  first  quadrant  will  be  positive.  But  the  tan- 
gent of  the  arc  AM',  greater  than  90°,  is  determined  by  the 
intersection  of  the  two  lines  M'C  and  AT.  These  lines,  how- 
ever, do  not  meet  in  the  direction  AT  ;  but  they  meet  in  the 
opposite  direction  AV.  But  since  the  tangents  estimated  in  the 
direction  AT  are  positive,  those  estimated  in  the  direction  AV 
must  be  negative :  therefore,  the  tangents  of  all  arcs  which  ter- 
minate in  the  second  quadrant  will  he  negative. 

When  the  point  M'  reaches  the  point  B  the  tangent  AV  will 
become  equal  to  zero :  that  is, 

tang  1 80° =0. 

When  the  point  M'  passes  the  point  B,  and  comes  into  the 
position  N',  the  tangent  of  the  arc  ADN'will  be  the  line  AT : 


PLANE  TRIGONOMETRY.  213 

hence,  the  tangents  of  all  arcs  which  terminate  in  the  third  quad- 
rant are  positive. 

At  E  the  tangent  becomes  infinite :  that  is, 
tang  270°  =  00. 

When  the  point  has  passed  along  into  the  fourth  quadrant 
to  N,  the  tangent  of  the  arc  ADN'N  will  be  the  line  AV :  hence, 
the  tangents  of  all  arcs  which  terminate  in  the  fourth  quadrant 
are  negative. 

The  cotangents  are  estimated  from  the  line  ED.  Those  which 
lie  on  the  side  DS  are  regarded  as  positive,  and  those  which  lie 
on  the  side  DS'  as  negative.  Hence,  the  cotangents  are  posi- 
tive in  the  first  quadrant,  negative  in  the  second,  positive  in  the 
third,  and  negative  in  the  fourth.  When  the  point  M  is  at  B 
the  cotangent  is  infinite  ;  when  at  E  it  is  zero :  hence, 

cot  180°=— 00  ;  cot  270°  =  0. 
Let  q  stand  for  a  quadrant ;  then  the  following  table  will  show 
the  signs  of  the  trigonometrical  lines  in  the  different  quadrants. 

\q  2q  Sq  4q 

Sine  +  +  —  — 

Cosine  +  —  —  + 

Tangent  +  —       '      +  —    ^ 

Cotangent  -r  —  -f  — 

XIII.  In  trigonometry,  the  sines,  cosines,  &c.  of  arcs  or  an- 
gles greater  than  180°  do  not  require  to  be  considered  ;  the 
angles  of  triangles,  rectilineal  as  well  as  spherical,  and  the 
sides  of  the  latter,  being  always  comprehended  between  0  and 
180°.  But  in  various  applications  of  trigonometry,  there  is  fre- 
quently occasion  to  reason  about  arcs  greater  than  the  semi- 
circumference,  and  even  about  arcs  containing  several  circum- 
ferences. It  will  therefore  be  necessary  to  find  the  expression 
of  the  sines  and  cosines  of  those  arcs  whatever  be  their 
magnitude. 

We  generally  consider  the  arcs  as  positive  which  are  esti- 
mated from  A  in  the  direction  ADB,  and  then  those  arcs  must 
be  regarded  as  negative  which  are  estimated  in  the  contrary 
direction  AEB. 

We  observe,  in  the  first  place,  that  two  equal  arcs  AM,  AN 
with  contraiy  algebraic  signs,  have  equal  sines  MP,  PN,  with 
contrary  algebraic  signs ;  while  the  cosine  CP  is  the  same  for 
both. 

The  equal  tangents  AT,  AV,  as  well  as  the  equal  cotangents 
DS,  DS',  have  also  contrary  algebraic  signs.  Hence,  calling 
X  the  arc,  we  have  in  general, 

sin  ( — x)= — sin  x 
cos  ( — ^a;)=cos  x 
tang  ( — x)= — ^tanga; 
cot  ( — x)= — cot  a; 


214 


PLANE  TRIGONOMETRY. 


By  considering  the  arc  AM,  and  its  supplement  AM',  and 
recollecting  what  has  been  said,  we  readily  see  that, 
sin  (an  arc)  =  sin  (its  supplement) 
cos  (an  arc)= — cos  (its  supplement) 
tang  (an  arc)= — tang  (its  supplement) 
cot  (an  arc)  = — cot  (its  supplement). 
It  is  no  less  evident,  that 
if  one  or  several  circumfe- 
rences  were   added  to  any 
arc  AM,  it  would  still  termi- 
nate exactly  at  the  point  M, 
and  the  arc  thus  increased 
would  have  the  same  sine  as 
the  arc  AM  ;  hence  if  C  rep- 
resent    a    whole    circumfe- 
rence or  360°,  we  shall  have 
sin  x= sin  (C  +  a;)=sinx=sin 
(2C  +  a:),  &c. 
The  same  observation  is  ap- 
phcable  to  the   cosine,  tan- 
gent, &c. 

Hence  it  appears,  that  whatever  be  the  magnitude  of  x  the 
proposed  arc,  its  sine  may  always  be  expressed,  with  a  proper 
sign,  by  the  sine  of  an  arc  less  than  180°.  For,  in  the  first 
place,  we  may  subtract  360°  from  the  arc  x  as  often  as  they 
are  contained  in  it ;  and  y  being  the  remainder,  we  shall  have 
sin  a;=sin  y.  Then  xiy  is  greater  than  180°,  make  y  =  180°  -}-%, 
and  we  have  sin  y= — sin  z.  Thus  all  the  cases  are  reduced 
to  that  in  which  the  proposed  arc  is  less  than  180°  ;  and  since 
we  farther  have  sin  (90°  +  x)=sin  (90°-^x),  they  are  likewise 
ultimately  reducible  to  the  case,  in  which  the  proposed  arc  is 
between  zero  and  90°. 

XIV.  The  cosines  are  always  reducible  to  sines,  by  means 
of  the  formula  cos  A  =  sin  (90° — A)  ;  or  if  we  require  it,  by 
means  of  the  formula  cos  A = sin  (90°  + A):  and  thus,  if  we  can 
find  the  value  of  the  sines  in  all  possible  cases,  we  can  also  find 
that  of  the  cosines.  Besides,  as  has  already  been  shown,  that 
the  negative  cosines  are  separated  from  the  positive  cosines  by 
the  diameter  DE ;  all  the  arcs  whose  extremities  fall  on  the 
right  side  of  DE,  having  a  positive  cosine,  while  those  whose 
extremities  fall  on  the  left   have  a  negative  cosine. 

Thus  from  0°  to  90°  the  cosines  are  positive  ;  from  90°  to 
270°  they  are  negative  ;  from  270°  to  360°  they  again  become 
positive ;  and  after  a  whole  revolution  they  assume  the  same 
values  as  in  the  preceding  revolution,  for  cos  (360°+a:)=cosar. 


PLANE  TRIGONOMETRY. 


215 


From  these  explanations,  it  will  evidently  appear,  that  the 
sines  and  cosines  of  the  various  arcs  which  are  multiples  of  the 
quadrant  have  the  following  values : 


sm  0°=0 
sin  180° =0 
sin  360° =0 
sin  540° =0 
sin  720°  =  0 
&c. 


sm    90° =R 
sin  270°=— R 
sin  450° =R 
sin  630°=— R 
sin  810°=R 
&c. 


cos      0°=R 
cos  180°=— R 
cos  360° =R 
cos  540°=— R 
cos  720° =R 
iStc. 


cos  90° =0 
cos  270° =0 
cos  450° =0 
cos  630° =0 
cos  810°=0 
&c. 


And  generally,  k  designating  any  whole  number  we  shall 
have 

sin  2^.90°=0,  cos  (2A:+1)  .  90°=0, 

sin  (4^+1)  .  90° =R,  cos  4^  .  90° =R, 

sin  (4A:— 1)  .  90°=— R,  cos  (4A;  +  2)  .  90°=— R. 
What  we  have  just  said  concerning  the  sines  and  cosines 
renders  it  unnecessary  for  us  to  enter  into  any  particular  de- 
tail respecting  the  tangents,  cotangents,  &c.  of  arcs  greater 
than  180°  ;  the  value  of  these  quantities  are  always  easily  de- 
duced from  those  of  the  sines  and  cosines  of  the  same  arcs : 
as  we  shall  see  by  the  formulas,  which  we  now  proceed  to 
explain. 


THEOREMS   AND  FORMULAS  RELATING  TO  SINES,  COSINES, 
TANGENTS,  &c. 

XV.  The  sine  of  an  arc  is  half  the  chord  which  subtends  a 

double  arc. 


For  the  radius  CA,  perpen- 
dicular to  the  chord  MN,  bi- 
sects this  chord,  and  likewise 
the  arc  MAN  ;  hence  MP,  the 
sine  of  the  arc  MA,  is  half  the 
chord  MN  which  subtends 
the  arc  MAN,  the  double  of 
MA. 

The  chord  which  subtends 
the  sixth  part  of  the  circum- 
ference is  equal  to  the  radius  ; 
hence 

^^^°-or  sin  30°=4R, 


sm 


12 


S' 


D 


\s 

r/y"'^^ 

a^ 

P 

^ 

P'    \ 

/ 

A 

Jb  — 

\ 

\ 

N' 

t^ 

V 

E 


m  other  words,  the  sine  of  a  third  part  of  the  right  angle  is 
equal  to  the  half  of  the  radius 


21Q 


PLANE  TRIGONOMETRY* 


S' 


D 


S 


N 

sy^ 

Q~^ 

f 

P'  \ 

/ 

^ 

^ 

y^ 

i 

N' 

\ 

R      J^ 

V 

E 


XVI.  The  square  of  the  sine 
of  an  arc^  together  with  the 
square  of  the  cosine,  is  equal 
to  the  square  of  the  radius  ;  so 
that  in  general  terms  we  have 

siii2A4-cos2A=R2. 

This  property  results  im- 
mediately from  the  right-an- 
gled triangle  CMP,  in  which 
MP2+CP2=CM2. 

It  follows  that  when  the 
sine  of  an  arc  is  given,  its  co- 
sine may  be  found,  and  re- 
ciprocally, by  means  of  the 

formulas  cos  A = db  V  (R^— sin^A),  and  sin  A = di:  V  (R^— cos^A) . 
The  sign  of  these  formulas  is  +,  or  — ,  because  the  same  sine 
MP  answers  to  the  two  arcs  AM,  AM',  whose  cosines  CP,  CP', 
are  equal  and  have  contrary  signs ;  and  the  same  cosine  CP 
answers  to  the  two  arcs  AM,  AN,  whose  signs  MP,  PN,  are 
also  equal,  and  have  contrary  signs. 

Thus,  for  example,  having  found  sin  30°=iR,  we  may  de- 
duce from  itcos30°,orsin60°=N/(R2— iR2)=v/fR2rriRv/3. 

XVII.  The  sine  and  cosine  of  an  arc  A  being  given,  it  is  re- 
quired to  find  the  tangent,  secant,  cotangent,  and  cosecant  of  the 
same  arc. 

The  triangles  CPM,  CAT,  CDS,  being  similar,  we  have  the 
proportions  : 


CP  :  PM  :  :  CA  :  AT ;  or  cos  A  :  sin  A  : :  R  :  tang  A= 


Rsin  A 


CP  :  CM  :  :  CA  ;  CT :  or  cos  A  :  R  :  :  R  :  sec  A= 


PM  :  CP  r :  CD  :  DS  ;  or  sin  A  :  cos  A  :  :  R  :  cot  A- 


PM  :  CM  :  :  CD  :  CS ;  or  sin  A  :  R  : :  R  :  cosec  A- 


cos  A 
_R2_ 

cos  A 

RcosA 

sin  A 

R2 


sin  A 


which  are  the  four  formulas  required.  It  may  also  be  observed, 
that  the  two  last  formulas  might  be  deduced  from  the  first  two, 
by  simply  putting  90° — A  instead  of  A. 

From  these  formulas,  may  be  deduced  the  values,  with  their 
proper  signs,  of  the  tangents,  secants,  &c.  belonging  to  any 
arc  whose  sine  and  cosine  are  known  ;  and  since  the  progres- 
sive law  of  the  sines  and  cosines,  according  to  the  different 
arcs  to  which  they  relate,  has  been  developed  already,  it  is 
unnecessary  to  say  more  of  the  law  which  regulates  the  tan- 
gents and  secants. 


PLANE  TRIGONOMETRY.  217 

By  means  of  these  formulas,  several  results,  which  have 
already  been  obtained  concerning  the  trigonometrical  lines, 
may  be  confirmed.  If,  for  example,  we  make  A =90°,  we 
shall  have  sin  A=R,  cos  A=0  ;  and  consequently  tang  90°  = 

p  2 

— ,  an  expression  which  designates  an  infinite  quantity  ;  for, 

the  quotient  of  radius  divided  by  a  very  small  quantity,  is  very 
great,  and  increases  as  the  divisor  diminishes  ;  hence,  the  quo- 
tient of  the  radius  divided  by  zero  is  greater  than  any  finite 
quantity. 

The  tangent  being  equal  to  R ;  and  cotangent  to  R.-^- ; 

cos  sm 

it  follows  that  tangent  and  cotangent  will  both  be  positive 
when  the  sine  and  cosine  have  like  algebraic  signs,  and  both 
negative,  when  the  sine  and  cosine  have  contrary  algebraic 
signs.  Hence,  the  tangent  and  cotangent  have  the  same 
sign  in  the  diagonal  quadrants  :  that  is,  positive  in  the  1st  and 
3d,  and  negative  in  the  2d  and  4th  ;  results  agreeing  with  those 
of  Art.  XII. 

It  is  also  apparent,  from  the  above  formulas,  that  the  secant 
has  always  the  same  algebraic  sign  as  the  cosine,  and  the  co- 
secant the  same  as  the  sine.  Hence,  the  secant  is  positive  on 
the  right  of  the  vertical  diameter  DE,  and  negative  on  the  left 
of  it ;  the  cosecant  is  positive  above  the  diameter  BA,  and  neg- 
ative below  it :  that  is,  the  secant  is  positive  in  the  1st  and  4th 
quadrants,  and  negative  in  the  2d  and  3d  :  the  cosecant  is  posi- 
tive in  the  1st  and  2d,  and  neg8^tive  in  the  3d  and  4th. 

XVIII.  The.  formulas  of  the  preceding  Article,  combined 
with  each  other  and  with  the  equation  sin ''^A  + cos  ^ArzR^, 
furnish  some  others  worthy  of  attention. 

First    we    have    R^  +  tang^    A  =  R^  +  ^'  ^'f^  ^  =z 

cos-*  A 

RMBin'A  +  cos^A)^_R^.  ^^„^^  R=+tang^  A=sec=  A,  a 

COS  -A  cos-  A 

formula  which  might  be  immediately  deduced  from  the  right- 
angled  triangle  CAT.  By  these  formulas,  or  by  the  right-an- 
gled triangle  CDS,  we  have  also  R'^  +  cot^  Arzcosec^  A. 

Lastly,  by  taking  .the  product  of  the  two  formulas  tang  A=r 

RsinA        1       ,4     RcosA  u         *         a      ^  a     t»o 

— ,  and  cot  A=  — : ,  we  have  tang  Ax  cot  A=R-,  a 

cos  A'  smA 

TJ2  p2 

■^     formula  which  gives  cot  A=-7- Xj.^ndtangA= t—t- 

T>2 

.  We  likewise  have  cot  B= . 

^  tangB 

T      28 


218 


PLANE  TRIGONOMETRY. 


Hence  cot  A  :  cot  B  :  :  tang  B  :  tang  A  ;  that  is,  the  cotan- 
gents of  two  arcs  are  reciprocally  proportional  to  their  tangents. 
The  formula  cot  Ax  tang  A=R'^  might  be  deduced  imme- 
diately, by  comparing  the  similar  triangles  CAT,  CDS,  which 
give  AT  ;  CA  :  :  CD  :  DS,  or  tang  A  :  R  :  :  R  :  cot  A. 

XIX.  The  sines  and  cosines  of  two  arcs^  a  and  b,  being  given, 
it  is  required  to  find  the  sine  and  cosine  of  the  sum  or  difference 
of  these  arcs. 

Let  the  radius  AC=R,the  arc 
AB=a,  the  arc  BD=fe,  and  con- 
sequently ABD=:a  +  h.  From 
the  points  B  and  D,  let  fall  the 
perpendiculars  BE,  DF  upon  AC ; 
from  the  point  D,  draw  DI  per- 
pendicular to  BC  ;  lastly,  from 
the  point  I  draw  IK  perpendicu- 
lar,  and  IL  parallel  to,  AC.  F'        C    FX  KE    P 

The  similar  triangles  BCE,  ICK,  give  the  proportions, 

CB  :  CI  :  :  BE  :  IK,or  R  :  cos  6  :  :  sin«  :  ik='^"^^^'^- 


R 

cos  a  cos  h. 

R" 


CB  :  CI  :  :  CE  :  CK,  or  R  :  cos  6  : :  cos  a  :  CK=: 

The  triangles  DIL,  CBE,  having  their  sides  perpendicular, 
each  to  each,  are  similar,  and  give  the  proportions, 

CB  :  DI  :  :  CE  :  DL,  or  R  ;  sin  6  :  :  cos  a  :  DL= 


cos  a  sin  b. 


R 


CB  :  DI  :  :  BE  :  IL,  or  R  :  sin  b 


sm  a 


IL 


sm  a 


sin  b. 


R 


But  we  have 
IK+DL-DF: 
Hence 


sin  (a +  5),  and  CK— IL  =: CF=;Cos  (a+b). 
sin  a  cos  &  +  sin  i  cos  a 


sin  {a-\-b)=: 
cos  (a  4-?))=: 


R 

cos  a  cos  b — sin  a  sin  b. 
R 


The  values  of  sin  (a — b)  and  of  cos  (a — b)  might  be  easily 
deduced  from  these  two  formulas  ;  but  .they  may  be  found 
directly  by  the  same  figure.  For,  produce  the  sine  DI  till  it 
meets  the  circumference  at  M;  then  we  have  BM=BD=6, 
and  MI = ID  =  sin  b.  Through  the  point  M,  draw  MP  perpen- 
dicular, and  MN  parallel  to,  AC  :  since  MIr=DI,  we  have  MN 
=IL,  and  IN=DL.  But  we  have  IK— lN=MP=sin  (a—b), 
and  CK + MN = CP = cos  (a—b) ;  hence 


PLANE  TRIGONOMETRY.  219 

.    /       ,^     sin  a  cos  h — sin  h  cos  a 
sin  {a—h)  = : g 

,/    cos  a  cos  6  +  sin  «5  sin  h 

cos  (a — o)  = 

R 

These  are  the  formulas  which  it  was  required  to  find. 

The  preceding  demonstration  may  seem  defective  in  point 

of  generahty,  since,  in  the  figure  which  we  have  followed,  the 

arcs  a  and  6,  and  even  «  +  &,  are  supposed  to  be  less  than  90°. 

But  first  the  demonstration  is  easily  extended  to  the  case  in 

which  a  and  h  being  less  than  90°,  their  sum  a-\-h'\s  greater 

than  90°.     Then  the  point  F  would  fall  on  the  prolongation  of 

AC,  and  the  only  change  required  in  the  demonstration  would 

be  that  of  taking  cos  {a  +  h)  = — CF' ;  but  as  we  should,  at  the 

same  time,  have  CF'^FL' — CK',  it  would  still  follow  that  cos 

(a4-&)  =  CK' — FL',  or  R  cos  (a +  6)= cos  a  cos  h — sin  a  sin  h. 

And  whatever  be  the  values  of  the  arcs  a  and  h,  it  is  easily, 

shown  that  the  formulas  are  true :  hence  we  may  regard  them 

as  established  for  all  arcs.     We  will  repeat  and  number  the 

formulas  for  the  purpose  of  more  convenient  reference. 

.     .    ,  ,v      sin  a  cos  ft  +  sin  h  cos  a  .  '  ^ 
sm  (a-f  &)  =  - -^ (1.). 

.     ,     . ,  v"     sin  a  cos  h — sin  h  cos  a  .„  v 
sm  {a—h)=- (2.). 

, ,     cos  a  cos  h — sin  a  sin  6  ,    , 
cos  (a  +  6)= — : ^ (3.) 

-.      cos  a  cos  6+.sin  a  smh  .   .     • 
cos  {a—b)  = g^ (4.) 

XX.  If,  in  the  formulas  of  the  preceding  Article,  we  make 
b=a,  the  first  and  the  third  will  give 

.    ^       2  sin  «  cos  a         ^       cos^  a — sin^  a     2  cos^  a — R" 
sm  2«= p ,  cos  2a= ^ = ^ 

formulas  which  enable  us  to  find  the  sine  and  cosine  of  the 

double  arc,  when  we  know  the  sine  and  cosine  of  the  arc  itself. 

Reciprocally,  to  divide  a  given  arc  «,  into  two  equal  parts, 

let  us,  in  the  same  formulas,  put  \a  instead  of  a :  we  shall  have 

sin  ^^asini^cosjfl    ^^^  ^^cos^  i«-sin^  ig 
R  R 

Now,      cos^  \a  +  s\v?  ifl!=R^  and  cos^^a — sin^  \az=iR  cos  a, 
there  results  by  adding  and  subtracting 

cos2  ia=iR2+xR  cos  a,  and  ^m^\a=:^W — iR  cos  a ; 
whence 

sin  ia-=  n/  GR2— iR  cos  ti)=\s^2W—2R  Qo^a, 

cos  |a=  v/(iRHiR  cos  a)==iN/2R2+2Rcosa. 


220  PLANE  TRIGONOMETRY. 

If  we  put  2a  in  the  place  of  a,  we  shall  have, 


sin  a=  \^(iR2— iR  cos  2a)=iV2K^—2R  cos  2a. 


cos  a=\/(iR2  +  iR  cos2a)=:iV2RH2R  cos  2a. 
Making,  in  the  two  last  formulas,  a=r:45°,  gives  cos  2a=0,  and 

sin  45°=  ViR2^Rv/i;  ind  also,  cos  45°=V^^=RVl, 

Next,  make  a =22°  30',  which  gives  cos  2a=^B  v^i,  and  we  have 

sin  22°  30'=R  (Vi — i%^i,  and  cos  22°  30'=Rv^(^+iv^i). 

XXI.  If  we  multiply  together  formulas  (1.)  and  (2.)  Art. 
XIX.  and  substitute  for  cos^  a,  R^ — sin^  a,  and  for  cos^  6, 
R^ — sin^  b ;  we  shall  obtain,  after  reducing  and  dividing  by  R^ 
sin  (a  +  b)  sin  (a — b) = sin^  a — sin^  b  =  (sin  a  +  sin  b)  (sin  a — sin  b). 

or,  sin  (a — b)  :  sin  a — sin  b  :  :  sin  a  +  sin  ft  ;  sin  {a  +  b). 

XXII.  The  formulas  of  Art.  XIX.  furnish  a  great  number  of 
consequences  ;  among  which  it  will  be  enough  to  mention  those 
of  most  frequent  use.  By  adding  and  subtracting  we  obtain 
the  four  which  follow, 

2 

sin  (a + ft)  +  sin  (a — 6)=_-sin  a  cos  b. 

R 

2 

•    sin  (a-ffe) — sin  (a — ft)  =^  sin  ft  cos  a, 

R 

2 

.   cos  (a  +  ft)  +  cos  (a — ft)  =  ^|5-cos  a  cos  ft. 

R 

2    . 

cos  (a — ft) — cos  (a  +  ft)=-^sin  a  sin  ft. 


R' 

and  which  serve  to  change  a  product  of  several  sines  or  co- 
sines into  lineal'  sines  or  cosines,  that  is,  into  sines  and  cosines 
multiplied  only  by  constant  quantities. 

XXIII.  If  in  these  formulas  we  put  a+ft==i?, « — &=^>  which 
p-\-Q        V — q 
gives  a—-^,  ^~~2~ '  ^®  ^^^^^  ^^ 

2 

sinp  +  sin  ^=-p^i»  \{p-^<l)  cos  \{p—q)  (1.) 

2 
sin  i?— sin  ^=^sin  ^  {p—q)  cos  J  (p  +  q)  (2.) 

cos;)  +  cos  ^r  =  —-cos  \{p-^q)  cos  I  {p — q)  (3.) 
R 

cosg — cosp=^sin  i{p  +  q)  sin  J  (p — q)  (4.) 


PLANE  TRIGONOMETRY.  221 

If  we  make  q=Oy  we  shall  obtain, 

2  sin  i  p  cos  i  p 
sin  p  = ^^ ^-^ 

MX 

^  2  cos^  i  p 

R+cos;?=     ^  ^^ 

R — cos  j»= — ^^ —  :  hence 

R 

sin  ^  _  tang  ^p  R 
R+cos  p~      R      ~cot  Jp 

sin  ^  _  cot  i  p  _  R 
R — cos  p  R  tang  j^p  : 

formulas  which  are  often  employed  in  trigonometrical  calcula- 
tions for  reducing  two  terms  to  a  single  one. 

XXIV.  From  the  first  four  formulas  of  Art  XXIII.  and  the  first 

of  Art.  XX.,  dividing,  and  considering  that =  — M-i= 

^  ^        cos  a        R        cota 

we  derive  the  following : 

sin  j?  +  sin  q  ^ sin  -^  {p-{-q)  cos  ^  (p — q)  _  tang  ^  (p  +  q) 

sin  J9— sin  q     cos^  \p  +  q)  sin  J  {p — q)     tang  ^(pHq) 

sin  jp  +  sin  <y_sin  j^  (j^  +  ^)_tang^  (P  +  q) 

cosp  +  cos^     cos\{p-{-q)  R 

sin  ./>-!- sin  y  _  cos  j^  {p — q)  _  cot  ^  ( jp — ^) 

cos  ^ — cos^    sinj(j9— (/)  R 

sin  jo— sin  g_sin  ^  {p — q)  _tang  ^  {p — q) 

cosp  +  cosq     C0sj(^— ^j     .  R 

sin  ^ — sin  q  _  cos  J  {p  -\-  q)  _•  cot  i  (p  +  q) 

COS  g— -cos  ^     sin  i  {p  +  q)  R 

cos  ;?  +  cos  g  ^cos  I  (p  +  q)  cos  ^  ( jo — q)  _^  cot  |  (jp  -f  {?) 

cos  (7 — cos^     sin|^(jw  +  ^)'sin  J  (^— ^)     tang  J  (jo — q) 

sin  p  +  sin  q     2sin  ^  (p  +  q)  cos  4  (/? — q)     cos  ^  (jo — q) 

sin      (i?  +  ^)"~2sini(jo-F^)cosi(jo  +  5')~cos^(p  +  ^) 

sin  jp — sin  q     2sin^  (jo — q)  cos^  (p  +  q)     sin  ^  (jp — gr) 

sm     (jp  +  g)~2sin  i  (jp  +  5')  cos i  (p  +  q) ~sin  i  (p  +  ^)" 

Formulas  which  are  the  expression  of  so  many  theorems. 
From  the  first,  it  follows  that  the  sum  of  the  sines  of  two  arcs  is 
to  the  difference  of  these  sines,  as  the  tangent  of  half  the  sum  of 
the  arcs  is  to  the  tangent  of  half  their  difference. 

T* 


222  PLANE  TRIGONOMETRY. 

XXV.  In  order  likewise  to  develop  some  formulas  reUtiye 
to  tangents,  let  us  consider  the  expression 

tang  {a-\-b)= -^ — J^^  ^^  which  by  substituting  the  values 

of  sin  (a+b)  and  cos  (cf  +  ft),  we  shall  find 

/    .  i-\     R  (sin  a  cos  5  +  sin  6  cos  a) 

cos  a  cos  b — sm  o  sm  a 

,,              ,          .          cos  a  tane  a       ,    .      "    cos  b  tang  b 
Now  we  have  sm  a= ^         ,  and.sm  b= p — ^-  : 

substitute  these  values,  dividing  all  the  terms  by  cos  cf  cos  b; 

we  shall  have 

7    .  i.\     K^  (tang  a  +  tang  b) 
tang  {a  +  h)=      "^   ^     \        /  5 
^  W — tang  a  tang  b 

which  is  the  value  of  the  tangent  of  the  sum  of  two  arcs,  ex- 
pressed by  the  tangents  of  each  of  these  arcs.  For  the  tangent 
of  their  difference,  we  should  in  like  manner  find 
/      \\     R^  (tang  a-^tang  b) 
*^"§  (^~^)=RH- tang  «  tang  &.  ' 

Suppose  b=a;  for  the  duplication  of  the  arcs,  we  shall  have 
the  formula 

2  R2  tang  a 

^^"g2^=R^_tang^a- 
Suppose  b=2a;  for  their  triplication,  we  shall  have  the  for- 
mula 

.         tang3«^g;ft^"g^  +  ^^"g^/); 
&  R2_tang  a  tang  2  a 

in  which,  substituting  the  value  of  tang  2  a,  we  shall  have 
^  R'^— 3tang2«. 

XXVI.  Scholium,  The  radius  R  iDeing  entirely  arbitrai-y,  is 
generally  taken  equal  to  1,  in  which  case  it  does  not  appear  in 
the  trigonometrical  formulas.  For  example  the  expression  for 
the  tangent  of  twice  an  arc  when  R=l,  becomes, 

^        ^         2  tang  a 

tang  2  a= 5_— 

1 — tang^  a- 

If  we  have  an  analytical  formula  calculated  to  the  radius  of  1, 
and  wish  to  apply  it  to  another  circle  in  which  the  radius  is  R, 
we  must  multiply  each  term  by  such  a  power  of  R  as  will  make 
all  the  terms  homogenious :  that  is,  so  that  each  shall  contain  the 
same  number  of  literal  factors. 


PLANE  TRIGONOMETRY.  22S 


CONSTRUCTION  AND  DESCRIPTION  OF  THE  TABLES. 

XXVII.  If  the  radius  of  a  circle  is  taken  equal  to  1,  and  the 
lengths  of  the  lines  representing  the  sines,  cosines,  tangents, 
cotangents,  &c.  for  every  minute  of  the  quadrant  be  calculated, 
and  written  in  a  table,  this  would  be  a  table  of  natural  sines, 
cosines,  <&c. 

XXVIII.  If  such  a  table  were  known,  it  would  be  easjr  to 
calculate  a  table  of  sines,  &c.  to  any  other  radius ;  since,  in 
different  circles,  the  sines,  cosines,  &c.  of  arcs  containing  the 
same  number  of  degrees,  are  to  each  other  as  their  radii. 

XXIX.  If  the  trigonometrical  lines  themselves  were  used,  it 
would  be  necessary,  in  the  calculations,  to  perform  the  opera- 
tions of  multiplication  and  division.  To  avoid  so  tedious  a 
method  of  calculation,  we  use  the  logarithms  of  the  sines,  co- 
sines, &c. ;  so  that  the  tables  in  common  use  show  the  values 
of  the  logarithms  of  the  sines,  cosines,  tangents,  cotangents,  &c, 
for  each  degree  and  minute  of  the  quadrant,  calculated  to  a 
given  radius.  This  radius  is  10,000,000,000,  and  consequently 
its  logarithm  is  10. 

XXX.  Let  us  glance  for  a  moment  at  one  of  the  methods 
of  calculating  a  table  of  natural  sines. 

The  radius. of  a  circle  being  1,  the  circumference  is  known 
to  be  3.14159265358979.  This  being  divided  successively,  by 
180  and  60,  or  at  once  by  10800,  gives  .0002908882086657, 
for  the  arc  of  1  minute.  Of  so  small  an  arc  the  sine,  chord, 
and  arc,  differ  almost  imperceptibly  from  the  ratio  of  equality  ; 
so  that  the  first  ten  of  the  preceding  figures,  that  is,  .0002908882 
may  be  regarded  as  the  «ine  of  1' ;  and  in  fact  the  sine  given 
in  the  tables  which  run  to  seven  places  of  figures  is  .0002909. 
By  Art.  XVI.  we  have  for  any  arc,  cos=  \/(l — sin^).  This 
theorem  gives,  in  the  present  case,  cos  1'  =  . 9999999577.  Then 
by  Art.  XXIT.  we  shall  have 

2  cos  I'xsin  1'— sin  0'=:sin  2' =  .00058 17764 
.2  cos  I'xsin  2'~sin  l'=sin  3' =  .0008726646 
2  cos  r  X  sin  3'-— sin  2'  =  sin  4' =  .00 11635526 
2  cos  I'xsin  4'— sin  3'  =  sin  5' =  .00 14544407 
2  cos  r  X  sin  5'— sin  4'=sin  6' =.001 7453284 
&;c.  &c.  &LC. 

Thus  may  the  work  be  continued  to  any  extent,  the  whole 
diflficulty  consisting  in  the  multiplication  of  each  successive  re- 
sult by  the  quantity  2  cos  1'  =  1.9999999154. 


224  PLANE  TRIGONOMETRY. 

Or,  the  sines  of  1 '  and  2'  being  determined,  the  work  might 
be  continued  thus  (Art.  XXI.) : 

sin  r  :  sin  2' — sin  1' : :  sin  2'  +  sin  1' :  sin  3' 
sin  2' :  sin  3' — sin  1' : :  sin  3'  + sin  1' :  sin  4' 
sin  3' :  sin  4' — sin  T  : :  sin  4'  +  sin  1' :  sin  5' 
sin  4' :  sin  5' — sin  1' : :  sin  5'  + sin  1' :  sin  6' 
.&c.  &c.  &c. 

In  hke  manner,  the  computer  might  proceed  for  the  sines  of 
degrees,  &c.  thus  :  • 

sin  1°  :  sin  2°— sin  1°  : :  sin  2°  + sin  1°  :  sin  3° 
sin  2°  :  sin  3°-^sin  1°  ::  sin  3°  + sin  1°  :  sin  4° 
sin  3°  :  sin  4° — sin  T  : :  sin  4°  +  sin  1°  :  sin  5= 
&c.  &c.  &c. 

Above  45°  the  process  may  be  considerably  simpUfied  by 
the  theorem  for  the.  tangents  of  the  sums  and  differences  of 
arcs.  For,  when  the  radius  is  unity,  the  tangent  of  45°  is  also 
unity,  and  tan  (a  +  b)  will  be  denoted  thus  : 

tan(45°  +  6)=i+-^. 
^       ^    ^     1  — tan  b 

And  this,  again,  may  be  still  further  simplified  in  practice. 
Th§  secants  and  cosecants  may  be  found  from  the  cosines  and 
sines. 


TABLE  OF  LOGARITHMS. 

XXXI.  If  the  logarithms  of  all  the  numbers  between  r  and 
any  given  number,  be  calculated  and  arranged  in  a  tabular  form, 
such  table  is  called  a  table  of  logarithms.  The  table  annexed 
shows  the  logarithms  of  all  numbers  between  1  and  10,000. 

The  first  column,  on  the  left  of  each  page  of  the  table,  is  the 
column  of  numbers,  and  is  designated  by  the  letter  N ;  the  deci- 
mal part  of  the  logarithms  of  these  numbers  is  placed  directly 
opposite  them,  and  on  the  same  horizontal  line. 

The  characteristic  of  the  logarithm,  or  the  part  which  stands 
to  the  left  of  the  decimal  point,  is  always  known,  being  1  less  than 
the  places  of  integer  figures  in  the  given  number,  and  there- 
fore it  is  not  written  in  the  table  of  logarithms.  Thus,  for  all 
numbers  between  1  and  10,  the  characteristic  is  0 :  for  num- 
bers between  10  and  100  it  is  1,  between  100  and  1000  it  is 
2,  «Sz:c. 


PLANE  TRIGONOMETRY.  225 

PROBLEM. 
To  find  from  the  table  the  logarithm  of  any  number, 

.  CASE  I. 

When  the  number  is  less  than  100. 

Look  on  the  first  page  of  the  table  of  logarithms,  along  the 
columns  of  numbers  under  N,  until  the  number  is  found  ;  the 
number  directly  opposite  it,  in  the  column  designated  Log.,  is 
the  logarithm  sought.  ^  . 

CASE  II. 
When  the  number  is  greater  than  100,  and  less  than  10,000. 

Find,  in  the  column  of  numbers,  the  three  first  figures  of  the 
given  number.  Then,  pass  across  the  page,  in  a  horizontal 
line,  into  the  columns  marked  0,  1,2,  3,  4,  &c.,  until  you  come 
to  the  column  which  is  designated  by  the  fourth  figure  of  the 
given  number  :  to  the  four  figures  so  found,  tw^o  figures  taken 
from  the  column  marked  0,  are  to  be  prefixed.  If  the  four 
figures  found,  stand  opposite  to  a  row  of  six  figures  in  the  column 
marked  0,  the  two  figures  from  this  column,  which  are  to  be 
prefixed  to  the  four  before  found,  are  the  first  two  on  the  left 
hand  ;  but,  if  the  four  figures  stand  opposite  a  line  of  only  four 
figures,  you  are  then  to  ascend  the  column,  till  you  come  to  the 
line  of  six. figures :  the  two  figures  at  the  left  hand  are  to  be 
prefixed,  and  then  the  decimal  part  of  the  logarithm  is  obtained. 
To  this,  the  characteristic  of  the  logarithm  is  to  be  prefixed, 
which  is  always  one  less  than  the  places  of  integer  figures  in 
the  given  number.    Thus,  the  logarithm  of  1 122  is  3.049993. 

In  several  of  the  columns,  designated  0,  1,  2,  3,  &c.,  small 
dots  are  found.  Where  this  occurs,  a  cipher  must  be  written 
for  each  of  these  dots,  a«d  the  two  figures  which  are  to  be  pre- 
fixed, from  the  first  column,  are  then  found  in  the  horizontal 
line  directly  below.  Thus,  the  log.  of  2 188  is  3.340047,  the  two 
dots  being  changed  into  two  ciphers,  and  the  34  from  tlie 
column  0,  prefixed.  The  two  figures  from  the  colum  0,  must 
also  be  taken  from  the  line  below,  if  any  dots  shall  have  been 
passed  over,  in  passing  along  the  horizontal  line :  thus,  the  loga- 
rithm of  3098  is  3.491081,  the  49  from  the  column  0  being 
taken  from  the  line  310, 

29 


226  PLANE  TRIGONOMETRY. 


CASE   III. 

When  the  number  exceeds  10,000,  or  consists  of  five  or  more 
places  of  figures. 

Consider  all  the  figures  after  the  fomrth  from  the  left  hand, 
as  ciphers.  Find,  from  th.e  table,  the  logarithm  of  the  first  four 
places,  and  prefix  a  characteristic  which  shall  be  one  less  than 
the  number  of  places  including  the  ciphers.  Take  from  the  last 
column  on  the  right  of  the  page,  marked  D,  the  number  on  the 
same  horizontal  line  with  the  logarithm,  and  multiply  this  num- 
ber by  the  numbers  that*have  been  considered  as  ciphers: 
then,  cut  off  from  the  rlgflt  hand  as  many  places  for  decimals 
as  there  are  figures  in  the  multiplier,  and  add  the  product,  so 
obtained,  to  the  first  logarithm :  this  sum  will  be  the  logarithm 
sought. 

Let  it  be  required  to  find  the  logarithm  of  672887.  The  log. 
of  672800  is  found,  on  the  1 1th  page  of  the  table,  to  be  5.827886, 
after  prefixing  the  characteristic  5.  The  corresponding  num- 
ber in  the  column  D  is  65,  which  being  multiplied  by  87,  the 
figures  regarded  as  ciphers,  gives  5655 ;  then,  pointing  off  two 
places  for  decimals,  the  number  to  be  added  is  56.55.  This 
number  being  added  to  5.827886,  gives  5.827942  for  the  loga- 
rithm of  672887  ;  the  decimal  part  .55,  being  omitted. 

This  method  of  finding  the  logarithms  of  numbers,  from  the 
table,  supposes  that  the  logarithms  are  proportional  to  their 
respective  numbers,  which  is  not  rigorously  true.  In  the  exam- 
ple, the  logarithm  of  672800  is  5.827886  ;  the  logarithm  of 
672900,  a  number  greater  by  100,  5.827951  :  the  difference  of 
the  logarithms  is  65.  Now,  as  100,  the  difference  of  the  numbers, 
is  to  65,  the  difference  of  their  logarithms,  so  is  87,  the  diffe- 
rence between  the  given  number  .and  the  least  of  the  numbers 
used,  to  the  difference  of  their  logarithm's,  which  is  56.55  :  this 
difference  being  added  to  5.827886,  the  logarithm  of  the  less 
number,  gives  5.827942  for  the  logarithm  of  672887.  The  use 
of  the  column  of  differences  is  therefore  manifest. 

When,  however,  the  decimal  part  which  is  to  be  omitted  ex- 
ceeds .5,  we  come  nearer  to  the  true  result  by  increasing  the 
next  figure  to  the  left  by  1  ;  and  this  will  be  done  in  all  the 
calculations  which  follow.  Thus,  the  difference  to  be  added, 
was  nearer  57  than  56 ;  hence  it  would  have  been  more  exact 
to  have  added  the  former  number. 

The  logarithm  of  a  vulgar  fraction  is  6qual  to  the  loga- 
rithm of  the  numerator,  minus  the  logarithm  of  the  denom- 


PLANE  TRIGONOMETRY.  227 

inator.  The  logarithm  of  a  decimal  fraction  is  found,  hy  con- 
sidering it  as  a  whole  number j  and  then  prefixing  to  the  decimal 
part  of  its  logaiHthm  a  negative  chai-acteristic,  greater  hy  unity 
than  the  number  of  ciphers  between  the  decimal  point  and  the  first 
significant  place  of  figures.  Thus,  the  logarithm  of  .0412,  is 
2^614897. 


PROBLEM. 
To  find  from  the  table,  a  number  answering  to  a  given  logarithm. 

XXXII  Search,  in  the  column  of  logarithms,  for  the  decimal 
part  of  the  given  logarithm,  and  if  it  be  exactly  found,  set  down 
the  corresponding  number.  Then,  if  the  characteristic  of  the 
given  logarithm  be  positive,point  off,  from  the  left  of  the  number 
found,  one  place  more  for  whole  numbers  than  there  are  units 
in  the  characteristic  of  the  given  logarithm,  and  treat  the  other 
places  as  decimals ;  this  will  give  the  number  sought. 

If  the  characteristic  of  the  given  logarithm  be  0,  there  will 
be  one  place  of  whole  numbers  ;  if  it  be  — 1,  the  number  will 
be  entirely  decimal ;  if  it  be  — 2,  there  will  be  one  cipher  be- 
tween the  decimal  point  and  the  first  significant  figure  ;  if  it  be 
— 3,  there  will  be  two,  &c.  The  number  whose  logarithm  is 
1.492481  is  found  in  page  5,  and  is  31.08. 

But  if  the  decimal  part  of  the  logarithm  cannot  be  exactly 
found  in  the  table,  take  the  number  answering  to  the  nearest 
less  logarithm ;  take  also  from  the  table  the  corresponding  dif- 
ference in  the  column  D :  then,  subtract  this  less  logarithm  from 
the  given  logarithm  ;  and  having  annexed  a  sufficient  number 
of  ciphers  to  the  remainder,  divide  it  by  the  difference  taken 
from  the  column  D,  and  annex  the  quotient  to  the  number  an- 
swering to  the  less  logarithm :  this  gives  the  required  number, 
nearly.  This  rule,  like  the  one  for  finding  the  logarithm  of  a 
number  when  the  places  exceed  four,  supposes  the  numbers  to 
be  proportional  to*  their  corresponding  logarithms. 

Ex.  1.  Find  the  number  answering  to  the  logarithm  1.532708. 
Here, 

The  given  logarithm,  is  .         .         _         1.532708 

Next  less  logarithm  of  34,09,  is        -         -         1.532627 

Their  difference  is  --         -         -         -         81 

And  the  tabular  difference  is  128  :  hence 
128)  81.00  (63 
which  being  annexed  to  34,09,  gives  34.0963  for  the  number 
answering  to  the  logarithm  1.532708.  • 


liUS  PLANE  TRIGONOMETRY. 

Ex,  2.     Required  the  number  answering  to  tiie  logarithm 
3.233568. 

The  given  logarithm  is  3.233568 

The  next  less  tabular  logarithm  of  1712,  is  3.233504 

Diff.=  64 

Tab.  Diff.=253)  64.00  (^5 
Hence  the  number  sought  is  1712.25,  marking  four  places 
of  integers  for  the  characteristic  3. 


TABLE  OF  LOGARITHMIC  SINES. 

XXXIII.  In  this  table  are  arranged  the  logarithms  of  the 
numerical  values  of  the  sines,  cosines,  tangents,  and  cotangents, 
of  all  the  arcs  or  angles  of  the  quadrant,  divided  to  minutes, 
and  calculated  for  a  radius  of  10,000,000,000.  The  logarithm 
of  this  radius  is  10.  In  the  first  and  last  horizontal  line,  of  each 
page,  are  written  the  degrees  whose  logarithmic  sines,  &:c,  are 
expressed  on  the  page.  The  vertical  columns  on  the  left  and 
right,  are  columns  of  minutes. 


CASE  L 

To  find,  in  the  table,  the  logarithmic  sine,  cosine,  tangent,  or  co- 
tangent of  any  given  arc  or  angle. 

1.  If  the  angle  be  less  than  45°,  look  in  the  first  horizontal 
line  of  the  different  pages,  until  the  number  of  degrees  be 
found  ;  then  descend  along  the  column  of  minutes,  on  the  left 
of  the  page,  till  you  reach  the  number  showing  the  minutes  ; 
then  pass  along  the  horizontal  line  till  you  come  into  the  column 
designated,  sine,  cosine,  tangent,  or  cotangent,  as  the  case  may 
be  :  the  number  so  indicated,  is  the  logarithm  nought.  Thus,  the 
sine,  cosine,  tangent,  and  cotangent  of  19°  55',  are  found  on 
page  37,  opposite  55,  and  are,  respectively,  9.532312,  9.973215, 
9.559097,  10.440903. 

2.  If  the  angle  be  greater  than  45°,  search  along  the  bottom 
line  of  the  different  pages,  till  the  number  of  degrees  are  found  ; 
then  ascend  along  the  column  of  minutes,  on  the  right  hand 
side  of  the  page,  till  you  reach  the  number  expressing  the  mi- 
nutes ;  then  pass  along  the  horizontal  line  into  the  columns 
designated  tang.,  cotang.,  sine,  cosine,  as  the  case  may  be  ;  the 
number  so  pointed  out  is  the  logarithm  required. 


I 


PLANE  TRIGONOMETRY.  229 

It  will  be  seen,  that  the  column  designated  sine  at  the  top  of 
the  page,  is  designated  cosine  at  the  bottom  ;  the  one  desig- 
nated tang.,  by  cotang.,  and  the  one  designated  cotang.,  by 
tang. 

The  angle  found  by  taking  the  degrees  at  the  top  of  the  page, 
and  the  minutes  from  the  first  vertical  column  on  the  left,  is  the 
complement  of  the  angle,  found  by  taking  the  corresponding 
degrees  at  the  bottom  of  the  page,  and  the  minutes  traced  up 
in  the  right  hand  column  to  the  same  horizontal  line.  This 
being  apparent,  the  reason  is  manifest,  why  the  columns  desig- 
nated sine,  cosine,  tang.,  and  cotang.,  when  the  degrees  are 
pointed  out  at  the  top  of  the  page,  and  the  minutes  counted 
downwards,  ought  to  be  changed,  respectively,  into  cosine,  sine, 
cotang.,  and  tang.,  when  the  degrees  are  shown  at  the  bottom 
of  the  page,  and  the  minutes  counted  upwards. 

If  the  angle  be  greater  than  90°,  we  have  only  to  subtract  it 
from  180°,  and  take  the  sine,  cosine,  tangent,  or  cotangent  of 
the  remainder. 

The  secants  and  cosecants  are  omitted  in  the  table,  being 
easily  found  from  the  cosines  and  sines. 

For,    sec.  = ;    or,  taking  the  logarithms,  log.  sec.=r2 

COS. 

log.  R — log.  cos.=20-^log.  COS. ;  that  is,  the  logarithmic  secant 
is  found  by  suhstracting  the  logarithmic  cosine  from  20.     And 

R^  • 

cosec  .  = ,  or  log.  cosec.=2  log.  R — log.  sine  =20 — log. 

sine 

sine  ;  that  is,  the  logarithmic  cosecant  is  found  by  subtracting  the 

logarithmic  sine  from  20. 

It  has  been  shown  that  R^mtang.  x  cotang. ;  therefore,  2  log. 
R:i=log.  tang.  +  log.  cotang.;  or  20=log.  tang.  +  log.  cotang. 

The  column  of  the  table,  next  to  the  column  of  sines,  and 
on  the  right  of  it,  is  designated  by  the  letter  D.  This  column 
is  calculated  in  the  following  manner.  Opening  the  table  at 
any  page,  as  42,  the  sine  of  24°  is  found  to  be  9.609313 ;  of 
24°  r,  9.609597 :  their  difference  is  284 ;  this  being  divided  by 
60,  the  number  of  seconds  in  a  minute,  gives  4.73,  which  is 
entered  in  the  column  D,  omitting  the  decimal  point.  Now, 
supposing  the  increase  of  the  logarithmic  sine  to  be  propor- 
tional to  the  increase  of  the  arc,  and  it  is  nearly  so  for  60",  it 
follows,  that  473  (the  last  two  places  being  regarded  as  deci- 
mals) is  the  increase  of  the  sine  for  1".  Similarly,  if  the  arc 
be  24°  20',  the  increase  of  the  sine  for  1",  is  465,  the  last  two 
places  being  decimals.  The  same  remarks  are  equally  apph- 
cable  in  respect  of  the  column  D,  after  the  column  cosine,  and 
of  the  column  D,  between  the  tangents  and  cotangents.  The 
column  D,  between  the  tangents    and  cotangents,    answers 

U 


23a  PLANE  TRIGONOMETRY,  ^ 

to  either  of  these  columns ;  since  of  the  same  arc,  the  lug. 
tang.  +  log.  cotang— 20.  Therefore,  having  two  arcs,  a  and  6, 
log.  tang  6  + log.  cotang  6:=:log.  tang  a  +  log.  cotang  q,\  or, 
log.  tang  h — log.  tang  a=log.  cotang  h — log.  cotang  a. 

Now,  if  it  were  required  to  find  the  logarithmic  sine  of  an 
arc  expressed  in  degrees,  minutes,  and  seconds,  we  hatve  only 
to  find  the  degrees  and  minutes  as  before  ;  then  multiply  the 
corresponding  tabular  number  by  the  seconds,  cut  off  two  places 
to  the  right  hand  for  decimals,  and  then  add  the  product  to  the 
number  first  found,  for  the  sine  of  the  given  arc.  Thus,  if  we 
wish  the  sine  of  40°  26'  28". 

The  sine  40°  26'         .         -         .         -         9.811952 

Tabular  difference  =:=  247 

Number  of  seconds  rr:    28 


Product  =  Q9.16,  to  be  added     -=  69.16 


Gives  for  the  sine  of  40°  26'  28"  =:  9.812021.16 

The  tangent  of  an  arc,  in  which  there  are  seconds,  is  found 
in  a  manner  entirely  similar.  In  regard  to  the  cosine  and  co-' 
tangent,  it  must  be  remembered,  that  they  increase  while  the 
arcs  decrease,  and  decrease  while  the  arcs  are  increased,  con- 
sequently, the  proportional  numbers  found  for  the  seconds  must 
be  subtracted,  not  added. 

Ex,  To  find  the  cosine  3°  40'  40". 

Cosine  3°  40^  9.999110 

Tabular  difference  ==  13 

Number  of  seconds  =  40 


Product  zr:  5.20,  which  being  subtracted    i=    5.20 


Gives  for  the  cosine  of  3°  40'  40"      9.999104.80 


CASE  II. 

To  find  the  degrees,  minutes ,  and  seconds  answering  to  any  given 
logarithmic  sine,  cosine,  tangent,  or  cotangent. 

Search  in  the  table,  and  in  the  proper  column,  until  the  num- 
ber be  found  ;  the  degrees  are  shown  either  at  the  top  or  bot- 
tom of  the  page,  and  the  minutes  in  the  side  columns,  either  at 
the  left  or  right.  But  if  the  number  cannot  be  exactly  found  in 
the  table,  take  the  degrees  and  minutes  answering  to  the  nearest 
less  logarithm,  the  logarithm  itself,  and  also  the  corresponding 
tabular  difference.     Subtract  the  logarithm  taken,  from  the 


PLANE  TRIGONOMETRY.  23 1 

given  logarithm,  annex  two  ciphers,  and  then  divide  the  re- 
mainder by  the  tabulai*  difference  :  the  quotient  is  seconds,  and 
is  to  be  connected  with  the  degrees  and  minutes  before  found  ; 
to  be  added  for  the  sine  and  tangent,  and  subtracted  for  the 
cosine  and  cotangent. 

Ex.  1.  To  find  the  arc  answering  to  the  sine       9.880054 
Sine  49°  20',  next  less  in  the  table,  9.879963 


Tab.  Diff.  181)9100(50" 

Hence  the  arc  49°  20'  50"  corresponds  to  the  given  sine 
9.880054. 

Ex.  2.  To  find  tJie  arc  corresponding  to  cotang.  10.008688f. 

Cotang  44°  26',  next  less  in  the  table  10,008591 

Tab.  Dift  421)9700(23" 

Hence,  44°  26' — 23" =44°  25'  37"  is  the  arc  corresponding 
to  the  given  cotangent  10.008688. 


r         0 


PRINCIPLES  FOR  THE  SOLUTION  OF  RECTIUNEAL  TRI- 
ANGLES. 

THEOREM  I. 

In  every  right  angled  triangle,  radius  is  to  the  sine  ^f  either 
of  the  acute  angles,  as  the  hypothenuse  to  the  opposite  side  . 
and  raditis  is  to  the  cosine  of  either  qfihe  acute  ^mgks,  as 
the  hypothenuse  to  the  adjacent  side. 

Let  ABC  be  the  proposed  tri-  ^ 

angle,  right-angled  at  A :  from 

the  point  C  as  a  centre,  with  a  v      ^y^ 

radius  CD  equal  to  the  radius  of  ^fT^ 

the  tables,  describe  the  arc  DE,  ^y^  \ 

which  will  measure  the  angle  C  ■;  ^^  \ 

on  CD  let  fall  the  perpendicular  r\^^ 1  \ 

EF,  which  will  be  the  sine  of  the   ^  :B*D  -flL 

angle  C,  and  CF  will  be  its  co- 
sine.   The  triangles  CBA,  CEF,  are  similar,  and  give  the  pro- 
portion, 

CE  :  EF  :  :  CB  :  BA :  hence 

H  :«inC  :  :  BC  :  BA. 


232 


PLANE  TRIGONOMETRY. 


■M'  -f. 


But  we  also  have,       -'^  «^ 

CE  :  CF  :  ;  CB  :  CA  :  hence 
R  :  cos  C  :  :  CB  ;  CA. 

Cor.     If  the  radius  R=l,  we  shall  have, 

AB=CB  sin  C,  and  CA  =  CB  cos  C. 

Hence,  in  every  right  angled  triangle,  the  perpendicular  is  equal 
to  the  hypothenuse  multiplied  by  the  sine  of  the  angle  at  the  base  ; 
and  the  base  is  equal  to  the  hypothenuse  multiplied,  by  the  cosine 
of  the  angle  at  the  base ;  the  radius  being  equal  to  unity. 


THEOREM  II. 


r^*> 


%im< 


In  every  right  angled  triangle,  radius  is  to  the  tangent  of  ei- 
ther of  the  acute  angles,  as  the  side  adjacent  to  the  side  op- 
posite. 

Let  CAB  be  the  proposed  tri- " 
angle. 

With  any  radius,  as  CD,  de- 
scribe the  arc  DE,  and  draw  the 
tangent  DG. 

From  the  similar  triangles 
CDG,CAB,  we  shall  have, 

CD  :  DG  ::  CA  :  AB  :  hence,    ^ 

R  :  tang  C  : :  CA  :  AB.  ; 

Cor.  1.     If  the  radius  R=l, 

AB=CA  tang  C. 
Hence,  the  perpendicular  of  a  right  angled  triangle  is  equal  to 
the  base  multiplied  by  the  tangent  of  the  angle  at  the  base,  the 
radius  being  unity. 

Cor.  2.  Since  the  tangent  of  an  arc  is  equal  to  the  cotangent 
of  its  complement  (Art.  VI.),  the  cotangent  of  B  may  be  sub- 
stituted in  the  proportion  for  tang  C,  which  will  give 
R  :  cot  B  ; :  CA :  AB. 


THEOREM  III. 


In  every  rectilineal  triangle,  the  sines  of  the  angles  are  to  each 
other  as  the  opposite  sides. 


PLANE  TRIGONOMETRY. 


233 


Let  ABC  be  the  proposed  triangle  ;  AD 
the  perpendicular,  let  fail  from  the  vertex  A 
on  the  opposite  side  BC  :  there  may  be  two 
cases. 

First.    If  the  perpendicular  falls  within  -^ 
the  triangle  ABC,  the  right-angled  triangles 
ABD,  ACD,  will  give, 

R  :  sin  B  : :  AB  :  AD. 
R  :  sin  C  : :  AC  :  AD. 

In  these  two  propositions,  the  extremes  are  equal ;  hence, 
sin  C  ;  sin  B  : :  AB  :  AC. 

Secondly.  If  the  perpendicular  falls 
without  the  triangle  ABC,  the  right- 
angled  triangles  ABD,  ACD,  will  still 
give  the  proportions, 

R  :  sin  ABD  : :  AB  :  AD, 
R:sinC        : :  AC  :  AD; 

from  which  we  derive 

sin  C  :  sin  ABD  : :  AB  :  AC. 

But  the  angle  ABD  is  the  supplement  of  ABC,  or  B ;  hence 
siji  ABD = sin  B ;  hence  we  still  have 

sinC;sinB::AB:AC. 


THEOREM  ly. 


In  every  rectilineal  triangle,  the  cosine  of  either  of  the  angles  is 
equal  to  radius  multiplied  by  the  sum  of  the  squares  of  the  sides 
adjacent  to  the  angle,  minus  the  square  of  the  side  opposite^ 
divided  by  twice  the  rectangle  of  the  adjacent  sides. 

Let  ABC  be  a  triangle  :  then  will 

AB2+BC2— AC2  s 


cos  B=- 


2AB  X  BC. 


First.  If  the  perpendicular  falls  within 

the  triangle,  we  shall  have  AC2=:AB2+ 

BC^— 2BC  X  BD  (Book  IV.  Prop.  XII.); 

,          j.^.    AB2+BC2--AC2    ^  ,  .    ,,      .  , ,       ,   ,  ,  .      , 
hence  BD= ^^^ .    But  m  the  right-angled  triangle 


D 


ABD,  we  have 


2BC 


R 


cos  B  : :  AB 
U*     30 


BD; 


234  PLANE  TRIGONOMETRY. 

„    RxBD       ,        ...       ,        ,       .«^ 

hence,  cos  13=       ^    ,  or  by  substituting  the  value  of  BD, 

^    „    AB2+BC2— AC2 
cosB=.Rx      ^^^^^^ 

Secondly.    If  the  perpendicular  falls 
without  the  triangle,  we    shall   have 
AC2=:ABHBC2+2BCxBD;    hence 
^^    AC2— AB2— BC2 

^^=" 2BC • 

But  in  the  right-angled  triangle  BAD,  

RxBD  D     B  C 

we  still  have  cos  ABD= — ~^u~  '•>  ^^^  t^©  angle  ABD  being 

supplemental  to  ABC,  or  B,  we  have 

T5                Auri         RxBD 
cos  B= — cos  ABD=: -^^  . 

hence  by  substituting  the  value  of  BD,  we  shall  again  have 
^    „    AB2+BC2— AC^ 
cosB^Rx       ^ABxBC      ' 

Scholium.  Let  A,  B,  C,  be  the  three  angles  of  any  triangle  ; 
a,  hy  c,  the  sides  respectively  opposite  them :  by  the  theorem, 

we  shall  have  cos  B=R  x  - — ^ .     And  the  same  principle, 

when  applied  to  each  of  the  other  two  angles,  will,  in  like  man- 

&2^c2_^2  a^-^y^-^ 

ner  give  cos  A=R  x ^r '>  and  cos  C=R  x  — ^"T — ' 

Either  of  these  formulas  may  readily  be  reduced  to  one  in  which 
the  computation  can  be  made  by  logarithms. 

Recurring  to  the  formula  R^ — R  cos  A=2sin^  J  A  (Art. 
XXIII.),  or  2sin^^A=R^ — RcosA,  and  substituting  for  cosA, 
we  shall  have 

2sinHA=:R^-R^x^^^^' 

_R^x26c-R^(&Hc^-~-a^  a^-^^-~c^-i-26c 

26c  ^  26c 

=  R.x±:|=i)Wx(?±^-4M^=^).    Hence 
26c  26c 

sin  JA=Rv/(fcth^g^±f=^). 

For  the  sake  of  brevity,  put 

J  (a4-6+c)=j9,  or  fl+6+c=2j9;  we  have  a +  6 — c=2p — ^2c, 

a  +  c — h^s=2p'^2b ;  hence 


PLANE  TRIGONOMETRY.  235 


THEOREM  V. 

In  every  rectilineal  triangle,  the  sum  of  two  sides  is  to  their  diffe- 
rence as  the  tangent  of  half  the  sum  of  the  angles  opposite  those 
sides,  to  the  tangent  of  half  their  difference. 

For,  AB  :  BC  :  :  sin  C  :  sin  A  (Theo- 
rem III,).  Hence,  AB  +  BC  :  AB— BC 
: :  sin  C  +  sin  A  ;  sin  C — sin  A.     But  sin 

sinC  +  sin  A :  sin  C — sin  A  : :  tang — - —  ; 
tang  ^^  (Art.  XXIV.)  ;  hence, 

AB+BC  :  AB— BC  :  :  tang  ^±^  :  tang  "^,   which  is 

the  property  we  had  to  demonstrate. 

With  the  aid  of  these  five  theorems  we  can  solve  all  the 
cases  of  rectilineal  trigonometry. 

Scholium.  The  required  part  should  always  be  found  from 
the  given  parts ;  so  that  if  an  error  is  made  in  any  part  of  the 
work,  it  may  not  affect  the  correctness  of  that  which  follows. 


SOLUTION  OF  RECTILINEAL  TRIANGLES  BY  MEANS  OF 
LOGARITHMS. 

It  has  already  been  remarked,  that  in  order  to  abridge  the 
calculations  which  are  necessary  to  find  the  unknown  parts  of 
a  triangle,  we  use  the  logarithms  of  the  parts  instead  of  the 
parts  themselves. 

Since  the  addition  of  logarithms  answers  to  the  multiplica- 
tion of  their  corresponding  numbers,  and  their  subtraction  to 
the  division  of  their  numbers  ;  it  follows,  that  the  logarithm  of 
the  fourth  term  of  a  proportion  will  be  equal  to  the  sum  of 
the  logarithms  of  the  second  and  third  terms,  diminished  by 
the  logarithm  of  the  first  term. 

Instead,  however,  of  subtracting  the  logarithm  of  the  first 
term  from  the  sum  of  the  logarithms  of  the  second  and  third 
terms,  it  is  more  convenient  to  use  the  arithmetical  complement 
of  the  first  term. 

The  arithmetical  complement  of  a  logarithm  is  the  number 
which  remains  after  subtracting  the  logarithm  from  10.  Thus 
10—9.274687  =  0.725313:  hence,  0.725313  is  the  arithmetical 
complement  of  9,274687, 


236  PLANE  TRIGONOMETRY. 

It  is  now  to  be  shown  that,  the  difference  between  two  loga- 
rithms is  truly  found,  by  adding  to  the  first  logarithm  the  arith- 
metical complement  of  the  logarithm  to  be  subtracted,  and  dimin- 
ishing their  sum  by  10. 

Let  a  =  the  first  logarithm. 

b  =  the  logarithm  to  be  subtracted. 

c  =  10 — &=the  arithmetical  complement  o(b. 

Now,  the  difference  between  the  two  logarithms  will  be 
expressed  by  a — b.  But  from  the  equation  c=10 — b,  we  have 
c — 10= — b  ;  hence  if  we  substitute  for  — b  its  value,  we  shall 
have 

a — b=a-\-c — 10, 
which  agrees  with  the  enunciation. 

When  we  wish  the  arithmetical  complement  of  a  logarithm, 
we  may  write  it  directly  from  the  tables,  by  subtracting  the 
left  hand  figure  from  9,  then  proceeding  to  the  right,  subtract 
each  figure  from  9,  till  we  reach  the  last  significant  figure,  which 
must  be  taken  from  10 :  this  will  be  the  same  as  taking  the 
logarithm  from  10. 

Ex.  From  3.274107  take  2.104729. 

Common  method.  By  ar.-comp. 

3.274107  3.274107 

2.104729  ar.-comp.  7.895271 

DifF.    1.169378  sum    1.169378  after    re- 

jecting the  10. 

We  therefore  have,  for  all  the  proportions  of  trigonometry, 
the  following 

RULE. 

Add  together  the  arithmetical  complement  of  the  logarithm  of  the 
the  first  term,  the  logarithm  of  the  second  term,  and  the  loga- 
rithm of  the  tliirdterm,  and  their  sum  after  rejecting  10,  will 
be  the  logarithm  of  the  fourth  term.  And  if  any  expression 
occurs  in  which  the  arithmetical  complement  is  twice  used,  20 
must  be  rejected  from  the  sum. 


PLANE  TRIGONOMETRY.  237 


SOLUTION  OF  RIGHT  ANGLED  TRIANGLES. 

Let  A  be  the  right  angle  of  the  proposed 
right  angled  triangle,  B  and  C  the  other  two 
angles ;  let  a  be  the  hypothenuse,  h  the  side 
opposite  the  angle  B,  c  the  side  opposite  the 
angle  C.  Here  we  must  consider  that  the  ^^  c" 
two  angles  C  and  B  are  complements  of  each  other  ;  and  that 
consequently,  according  to  the  different  cases,  we  are  entitled 
to  assume  sin  C=cos  B,  sin  B=cos  C,  and  likewise  tang  B= 
cot  C,  tang  C=cot  B.  This  being  fixed,  the  unknown  parts 
of  a  right  angled  triangle  may  be  found  by  the  first  two  theo- 
rems ;  or  if  two  of  the  sides  are  given,  by  means  of  the  pro- 
perty, that  the  square  of  the  hypothenuse  is  equal  to  the  sum 
of  the  squares  of  the  other  two  sides. 

EXAMPLES. 

Ex.  1.  In  the  right  angled  triangle  BCA,  there  are  given  the 
hypothenuse  «=250,  and  the  side  6=240 ;  required  the  other 
parts. 

R  :  sin  B  :  :  a  :  6  (Theorem  I.), 
or,  «  :  6  ;  :  R  :  sin  B. 

When  logarithms  are  used,  it  is  most  convenient  to  write  the 
proportion  thus. 

As  hyp.  a      -     250      -      ar.-comp.     log.     -     7.602060 

To  side  6      -     240 2.380211 

So  is      R      -----------  10.000000 

To  sin  B      -      73°  44'  23"  (after  rejecting  10)    9.982271 

But  the  angle  C=90°— B=90°— 73°  44' 23"=  16°  15'  37". 
or,  C  might  be  found  by  the  proportion, 

As  hyp.  a  -     250     -     ar.-comp.  log.     -     7.G02060 

To  side  6  -     240 2.380211 

So  is      R -  10.000000 

To  cos  C  -       16°  15'  37"     ------     9.982271 

To  find  the  side  c,  we  say. 

As  R        -         -         ar.  comp.  log.  -  0.000000 

To  tang.  C  16°  15' 37"      -  -         -  -  9.464889 

So  is  side  6  240         -         -  -  -  2.380211 

To  side      c  70.0003           -  -  -  1.845100 


238  PLANE  TRIGONOMETRY. 

Or  the  side  c  might  be  found  from  the  equation 

For,  c2=e{2-~i2=(^_,,  J)  X  (a—ft)  : 

hence,  2  log.  c=log.  («  +  ?>)  + log.  (a — 6),  or 

log.  c=Jlog.  (aH-6)+J]og.  (a— !^) 
«  + 6=250 +  240=490        log.  2.690196 

a— 6=250— 240=10     -         -         1.000000 

2)  3.690196 
Log.  c      70      -         -         -         -         -         -         1.845098 

^  JBa;.  2.  In  the  right  angled  triangle  BCA,  there  are  given, 
side 6=384  yards,  and  the  angle  B=53°  8' :  required  the  other 
parts. 

To  find  the  third  side  c. 
R  :  tang  B  :  :  c  :  6  (Theorem  II.) 
or,  tang  B  :  R  :  :  6  :  c.     Hence, 

As  tang     B  53°  8'  ar.-comp.         log.         9.875010 

Is  to  R  10.000000 

So  is  side  6  384 2.584331 


To  side  c  287.965         ....  2.459341 

Note.  When  the  logarithm  whose  arithmetical  complement 
is  to  be  used,  exceeds  10,  take  the  arithmetical  complement 
with  reference  to  20  and  reject  20  from  the  sum. 

To  find  the  hypothenuse  a, 
R  :  sin  B  :  :  a  :  6    (Theorem  I.).     Hence, 

As  sin  B  53°  8'         ar.  comp.         log.  0.096892 

Is  to  R              10.000000 

^  ^^    So  is  side  6  384         -         -         -         .  2.584331 

To  hyp.     a  479.979           -         -         -  2.681223 


Ex,  3.  In  the  right  angled  triangle  BAG,  there  are  given, 
side  c=195,  angle  B=47°  55', 
required  the  other  parts. 

Arts.  Angle  C=42°  05',  a=290.953,  6=215.937. 

SOLUTION  OF  RECTILINEAL  TRIANGLES  IN  GENERAL. 

Let  A,  B,  C  be  the  three  angles  of  a  proposed  rectilineal  tri- 
angle ;  a,  6,  c,  the  sides  which  are  respectively  opposite  them  ; 
the  different  problems  which  may  occur  in  determining  three  of 
these  quantities  by  means  of  the  other  three,  will  all  be  redu- 
cible to  the  four  following  cases. 


PLANE  TRIGONOMETRY. 


239 


CASE  I. 

Given  a  side  and  two  angles  of  a  triangle,  to  find  the  remaining 

parts. 

First,  subtract  the  sum  of  the  two  angles  from  two  right  an- 
gles, the  remainder  will  be  the  third  angle.  The  remaining 
sides  can  then  be  found  by  Theorem  III. 

I.  In  the  triangle  ABC,  there  are  given  the  angle  A =58°  07', 
the  angle  B=22°  37',  and  the  side  c=408  yards:  required  the 
remaining  angle  and  the  two  other  sides. 

To  the  angle  A          -         -         -         -         -         =58°  07' 
Add  the  angle  B =22°  37' 

Their  sum  -         -         -         -         -         .         =80°  44' 

taken  from  180°  leaves  the  angle  C  -         =99°  16'. 

This  angle  being  greater  than  90°  its  sine  is  found  by  taking 
that  of  its  supplement  80°  44'. 


As  sine  C 
Is  to  sine  A 
So  is  side  c 

So  side  a 


As  sine  C 
Is  to  sine  B 
So  is  side  c 
To  side  b 


To  find  the  side  a. 

99°  16'         ar.-comp.  log. 

58°  07'         .         .         -  - 

408 

351.024         -         .         .  - 

To  find  the  side  b. 

99°  16'         ar.-comp.  log. 

22°  37'         -         -         .  . 

408     .         -         -         -  - 
158.976        -         -         - 


0.005705 
9.928972 
2.610660 

2.545367 


0.005705 
9.584968 
2.610660 
2.201333 


2.  In  a  triangle  ABC,  there  are  given  the  angle  A =38°  25' 
B=57°  42',  and  the  side  C=400:  required  the  remaining 
parts. 

Ans,  Angle  C=83°  53',  side  a =249.974,  side  ^>=:340.04. 


CAS?  II. 


Criven  two  sides  of  a  triangle,  and  an  angle  opposite  one  of  them, 
to  find  the  third  side  and  the  two  remaining  angles. 


240 


PLANE  TRIGONOMETRY. 


7.931814 
2.334454 
9.584968 
9.851236 


1.  In  the  triangle  ABC,  there 
are  given  side  AC =216,  BC  = 
117,  and  the  angle  A =22°  37', 
to  find  the  remaining  parts. 

Describe  the  triangles  ACB, 
ACB',  as  in  Prob.  XI.  Book  III. 

Then  find  the  angle  B  by- 
Theorem  III. 

As  side  B'C  or  BC  117  ar.-comp.     log. 

Is  to  side  AC  216 

So  is  sine  A  22°  37' 

To  sine  B'  45°  13'  55"  or  ABC  134°  46' 05" 

Add  to  each  A    22°  37'  00"  22°  37'  00" 

Take  their  sum    67°  50'  55"  157°  23'  05" 

From  180°  00' 00"  180°  00' 00" 

Rem.     ACB'     112°  09' 05"     ACB=22°  36'  55" 

To  find  the  side  AB  or  AB'. 

As  sine  A  22°  37'        ar.-comp.      log. 

Is  to  sine  ACB'  112°  09' 05"      -^  ,,  i^ 

So  is  side  B'C    117         -       ,  - 

To  side  AB'     281.785         -  -         -        

The  ambiguity  in  this,  and  similar  examples,  arises  in  con- 
sequence of  the  first  proportion  being  true  for  both  the  trian- 
gles ACB,  ACB'.  As  long  as  the  two  triangles  exist,  the  am- 
biguity will  continue.  But  if  the  side  CB,  opposite  the  given 
angle,  be  greater  than  AC,  the  arc  BB'  will  cut  the  line  ABB', 
on  the  same  side  of  the  point  A,  but  in  one  point,  and  then 
there  will  be  but  one  triangle  answering  the  conditions. 

If  the  side  CB  be  equal  to  the  perpendicular  Cd,  the  arc 
BB'  will  be  tangent  to  ABB',  and  in  this  case  also,  there  will 
be  but  one  triangle.  When  CB  is  less  than  the  perpendicular 
Cd,  the  arc  BB'  will  not  intersect  the  base  ABB',  and  in  that 
case  there  will  be  no  triangle,  or  the  conditions  are  impossible. 

2.  Given  two  sides  of  a  triangle  50  and  40  respectively,  and 
the  angle  opposite  the  latter  equal  to  32°  :  required  the  remain- 
ing parts  of  the  triangle. 

Ans,  If  the  angle  opposite  the  side  50  be  acute,  it  is  equal 
to  41°  28'  59",  the  third  angle  is  then  equal  to  106°  31'  01",  and 
the  third  side  to  72.368.  If  the  angle  opposite  the  side  50  be 
obtuse,  it  is  equal  to  138°  31'  01",  the  third  angle  to  9°  28'  59", 
and  the  remaining  side  to  12.436. 


0.415032 
9.966700 
2.068186 
2.449918 


PLANE  TRIGONOMETRY.  241 


CASE  III. 

Given  two  sides  of  a  triangle^  with  their  included  angle,  to  find 
the  third  side  and  the  two  remaining  angles. 

Let  ABC  be  a  triangle,  B  the  given 
angle,  and  c  and  a  the  given  sides. 

Knowing  the  angle  B,  v^^e  shall  like- 
wise know  the  sum  of  the  other  two  an- 
gles C  +  A=180°— B,  and  their  half  sum 
1- (C  +  A)=90— JB.     We    shall    next    A  h  'C 

compute  the  half  difference  of  these  two  angles  by  the  propor- 
tion (Theorem  V.), 

c-f  a  ;  c — a  : :  tang  J  (C  +  A)  or  cot  J  B  :  tang  J  (C — A,) 

in  which  we  consider  c>fl  and  consequently  C>A.  Having 
found  the  half  difference,  by  adding  it  to  the  half  sum 
\  (C  +  A),  we  shall  have  the  greater  angle  C  ;  and  by  subtract- 
ing it  from  the  half-sum,  we  shall  have  the  smaller  angle  A. 
For,  C  and  A  being  any  two  quantities,  we  have  always, 

C=rHC  +  A)+J(C— A) 
A=:ri(C+A)-4(C-A). 

Knowing  the  angles  C  and  A  to  find  the  third  side  h,  we  have 
the  proportion. 

sin  A  :  sin  B  : :  a  :  6 

Ex,  1.  In  the  triangle  ABC,  leta=450,  c=540,  and  the  in- 
cluded angle  Bur  80°  :  required  the  remaining  parts. 

c  +  air=990,c— a=90,  180°— B  =  100°  =  C  +  A. 

Asc  +  a           990             ar.-comp.         log.  7.004365 

Is  toe— a          90 1.954243 

So  is  tang  J  (C  +  A)  50°           .         -         -  10.076187 

To  tang  \  (C— A)  6°  11'             -         -         -  9.034795 

Hence,  50°  +  6°  ll'=:56°  ir=C  ;  and  50°— 6°  ir=:43°  49' 

=A. 

To  find  the  third  side  h. 

As  sine  A       43°  49'         ar.-comp.         log.         0.159672 

Is  to  sine  B     80° 9.993351 

So  is  side  a     450      -         -         -         -         -         2.653213 
To  side  h        640.082        ...         -         2.806236 
Ex.  2.     Given  two  sides  of  a  plane  triangle,  1686  and  960, 
and  their  included  angle  128°  04':  required  the  other  parts. 

^715.  Angles,  33°  34'  39 ",  18°  21'  21",  side  2400. 
X31 


242  PLANE  TRIGONOMETRY. 

CASE  IV. 

Given  the  three  sides  of  a  triangle,  to  find  the  angles. 
We  have  from  Theorem  IV.  the  formula, 

sin  i  A=R  V^(1S(£E5)    i„^hich 
p  represents  the  half  sum  of  the  three  sides.     Hence, 
srn^A=k{P-^\lP-'),    or 

2  log.  sin  JA=2  log.  R  +  log.  (p — 6)  + log.  {p--c) — log.  c — 
log.  h. 

Ex.  1.     In  a  triangle  ABC,  let  b=40,  c=34,  and  a^25  : 

required  the  angles. 

Here  jo= ^ =49.5,  jo — &=^9.5,  and  /?— c=I5.5. 

2  Log.  R -  20.000000 

log.  ip—b)     9.5 0.977724 

log.  (p^c)  15.5       -         -         -         -         -  1.190332 

— log.  c  34  ar.-comp.      -         -  8.468521 

— log.  h  40  ar.-comp.       -         -  8.397940 

2  sin  J  A 19.034517 

log.  sin  J  A  19°  12'  ^9"  -         -         -  9-517258 

Angle  A=38°  25'  18". 

In  a  similar  manner  we  find  the  angle  B=83°  53'  18"  and 
the  angle  C=57°  41'  24". 

Ex.  2.  What  are  the  angles  of  a  plane  triangle  whose  sides 
are,  a=60,  6=50,  and  c=40? 

^715.  41°  24'  34",  55°  46'  16"  and  82°  49'  10". 


APPLICATIONS. 

Suppose  the  height  of  a  building  AB  were  required,  the 
foot  of  it  being  accessible. 


PLANE  TRIGONOMETRY. 


243 


On  the  ground  which  we 
suppose  to  be  horizontal  or  very 
nearly  so,  measure  a  base  AD, 
neither  very  great  nor  very 
small  in  comparison  with  the 
altitude  AB ;  then  at  D  place 
the  foot  of  the  circle,  or  what- 
ever be  the  instrument,  with 
which  we  are  to  measure  the 
angle  BCE  formed  by  the  hori- 
zontal line  CE  parallel  to  AD,  A.  D 
and  by  the  visual  ray  direct  it  to  the  summit  of  the  building- 
Suppose  we  find  AD  or  CE  1=67.84  yards,  and  the  angle 
BCE=4lo  04' :  in  order  to  find  BE,  we  shall  have  to  solve 
the  right  angled  triangle  BCE,  in  which  the  angle  C  and  the 
adjacent  side  CE  are  known. 

To  find  the  side  EB. 

As  R ar.-comp.      -  0.000000 

Is  to  tang.  C  41°  04' 9.940183 

So  is  EC  67.84 1.831486 

ToEB  59.111 1.771669 


Hence,  EB=59.111  yards.  To  EB  add  the  height  of  the 
instrument,  which  we  will  suppose  to  be  1.12  yards,  we  shall 
then  have  the  required  height  AB=60.231  yards. 

If,  in  the  same  triangle  BCE  it  were  required  to  find  the 
hypothenuse,  form  the  proportion 

As  cos  C  41°  04'        ar.-comp.        -     -       log.     0.122660 

Is  to  R 10.000000 

So  is  CE    67.84     .-.--.---       1.831486 

To  CB        89.98     -     - 1.954146 


Note.  If  only  the  summit  B  of  the  building  or  place  whose 
height  is  required  were  visible,  we  should  determine  the  dis- 
tance BE  by  the  method  shown  in  the  following  example ; 
this  distance  and  the  given  angle  BCE  are  suflicient  for  solv- 
ing the  right  angled  triangle  BCE,  whose  side,  increased  by 
the  height  of  the  instrument,  will  be  the  height  required. 


244 


PLANE  TRIGONOMETRY 


2.  To  find  upon  the  ground 
the  distance  of  the  point  A 
from  an  inaccessible  object 
B,  we  must  measure  a  base 
AD,  and  the  two  adjacent 
angles  BAD,  ADB.  Sup- 
pose we  have  found  AD= 
588.45  yards,  BAD  =  103° 
55'  55",  and  BDA  =  36^  04'; 
we  shall  thence  get  the  third 
angle  ABD=40°  05",  and  to 
obtain  AB,  we  shall  form  the 
proportion 

As  sine  ABD  40°  05" 
Is  to  sin  BDA  36°  04' 
So  is  AD  588.45 

ToAB  -     -    538.943 


ar.-comp. 


log. 


0.191920 
9.769913 
2.769710 


2.731543 


If  for  another  inaccessible  object  C,  we  have  found  the  an- 
gles CAD =35°  15',  ADC  =  119°  32',  we  shall  in  like  manner 
find  the  distance  AC  =  1201.744  yards. 

3.  To  find  the  distance  between  two  inaccessible  objects  B 
and  C,  we  determine  AB  and  AC  as  in  the  last  example ;  we 
shall,  at  the  same  time,  have  the  included  angle  BAC=BAD — 
DAC.  Suppose  AB  has  been  found  equal  to  538.818  yards, 
AC  =  1201.744  yards,  and  the  angle  BAC=68°  40'  55";  to 
get  BC,  we  must  resolve  the  triangle  BAC,  in  which  are  known 
two  sides  and  the  included  angle. 

AsAC  +  AB     1740.562     ar.-comp.        log.-  6.759311 

Is  to  AC— AB   662.926 2.821465 

B-t-C 

So  is  tang.— --    55°  39' 32" 10.165449 


To  tang. 


B— C 


29°  08'  19" 9.746225 


Hence    -    -    - 
But  we  have    - 


B— C 


2 
B  +  C 


=29°  08'  19' 


=  55°  39'  32' 


Hence B     =84°  47'  51" 

and         C     =26°  31'  13" 


I 


PLANE  TRIGONOMETRY.  245 

Now,  to  find  the  distance  BC  make  the  proportion, 

As  sine  B  84°  47'  51"     ar.-comp.     -     log.     -  0.001793 

Is  to  sine  A  68°  40'  55 ' 9.969218 

So  is  AC  1201.744 8.079811 

To  BC  1124.145 3.050822 

4.  Wanting  to  know  the  distance  between  two  inaccessible 
objects  which  lie  in  a  direct  line  from  the  bottom  of  a  tower 
of  120  feet  in  height,  the  angles  of  depression  are  measured, 
and  found  to  be,  of  the  nearest,  57° ;  of  the  most  remote, 
25°  30'  I  required  the  distance  between  them. 

Ans.  173.656  feet. 

6.  In  order  to  find  the  distance  between  two  trees,  A  and 
B,  which  could  not  be  directly  measured  because  of  a  pool 
which  occupied  the  intermediate  space,  the  distance  of  a  third 
point  C  from  each,  was  measured,  viz.  CA=588  feet  and  CB 
=672  feet,  and  also  the  contained  angle  ACB=55°  40':  requi- 
red the  distance  AB. 

Ans.  592.967  feet. 

6.  Being  on  a  horizontal  plane,  and  wanting  to  ascertain 
the  height  of  a  tower,  standing  on  the  top  of  an  inaccessible 
hill,  there  were  measured,  the  angle  of  elevation  of  the  top  of 
the  hill  40°,  and  of  the  top  of  the  tower  51°  :  then  measuring 
in  a  direct  line  180  feet  farther  from  the  hill,  the  angle  of  ele- 
vation of  the  top  of  the  tower  was  33°  45' :  required  the  height 
of  the  tower. 

Ans.  83.9983  feet. 

7.  Wanting  to  know  the  horizontal  distance  between  two 
inaccessible  objects  A  and  B,  and  not  finding  any  station  from 
which  both  of  them  could  be  seen,  two  points  C  and  D,  were 
chosen,  at  a  distance  from  each  other  equal  to  200  yards,  from 
the  former  of  which  A  could  be  seen,  and  from  the  latter  B, 
and  at  each  of  the  points  C  and  D  a  staff  was  set  up.  From 
C  a  distance  CF  was  measured,  not  in  the  direction  DC,  equal 
to  200  yards,  and  from  D,  a  distance  DE  equal  to  200  yards, 
and  the  following  angles  were  taken,  viz.  AFC=83°  ACF=i 
54°  31',  ACD=53°  30',  BDC=156°  25',  BDE=54°  30',  and 
BED =88°  30' :  required  the  distance  AB. 

Ans.  345.46  yards. 

8.  From  a  station  P  there  can  be  seen  three  objects,  A,  B 
and  C,  whose  distances  from  each  other  are  known,  viz.  AB= 
800,  AC  =  600,  and  BC— 400  yards.  There  are  also  measured 
the  horizontal  angles,  A?C=33°  45',  BPC=22°  30'.  It  is  re- 
quired, from  these  data,  to  determine  the  three  distances  PA, 
PC  and  PB. 

Ans.  PA=710.193,  PC  =  1042.522,  PB  =  934.291  yards. 

X* 


246  SPHERICAL  TRIGONOMETRY. 


SPHERICAL  TRIGONOMETRY. 

I.  It  has  already  been  shown  that  a  spherical  triangle  is 
formed  by  the  arcs  of  three  great  circles-  intersecting  each  other 
on  the  surface  of  a  sphere,  (Book  IX.  Def  1).  Hence,  every 
spherical  triangle  has  six  parts :  the  sides  and  three  angles. 

Spherical  Trigonometry  explains  the  methods  of  determin- 
ing, by  calculation,  the  unknown  sides  and  angles  of  a  spheri- 
cal triangle  when  any  three  of  the  six  parts  are  given. 

II.  Any  two  parts  of  a  spherical  triangle  are  said  to  be  of 
the  same  species  when  they  are  both  less  or  both  greater  than 
90<* ;  and  they  are  of  different  species  when  one  is  less  and  the 
other  greater  than  90°. 

III.  Let  ABC  be  a  spherical 
triangle,  and  O  the  centre  of  the 
sphere.  Let  the  sides  of  the  tri- 
angle be  designated  by  letters 
corresponding  to  their  opposite 
angles:  that  is, the  side  opposite 
the  angle  A  by  «,  the  side  oppo- 
site B  by  6,  and  the  side  opposite 
C  by  c.  Then  the  angle  COB 
will  be  represented  by  a,  the  an- 
gle CO  A  by  h  and  the  angle 
BOA  by  c.  The  angles  of  the 
spherical  triangle  will  be  equal  to  the  angles  included  between 
the  planes  which  determine  its  sides   (Book  IX.  Prop.  VI.). 

From  any  point  A,  of  the  edge  OA,  draw  AD  perpendicular 
to  the  plane  COB.  From  D  draw  DH  perpendicular  to  OB, 
and  DK  perpendicular  to  OC ;  and  draw  AH  and  AK :  the 
last  lines  will  be  respectively  perpendicular  to  OB  and  OC, 
(Book  VI.  Prop.  VI.) 

The  angle  DHA  will  be  equal  to  the  angle  B  of  the  spheri- 
cal triangle,  and  the  angle  DKA  to  the  angle  C. 

The  two  right  angled  triangles  OKA,  ADK,  will  give  the 
proportions 

R  :  sin  AOK  : :  OA  :  AK,  or,  R  x  AK=  OA  sin  b. 
R  :  sin  AKD  : :  AK  :  AD,  or,  R  x  ADrrAK  sin  C. 

Hence,  R^  x  AD = AO  sin  h  sin  C,  by  substituting  for  AK  its 
value  taken  from  the  first  equation. 


SPHERICAL  TRIGONOMETRY.  247 

In  like  manner  the  triangles  AHO,  ADH,  right  angled  at 
H  and  D,  give 

R  :  sin  c  : :  AO  :  AH,  or  R  x  AH=:AO  sin  c 
R  :  sin  B  : :  AH  :  AD,  or  Rx  AD=AH  sin  B. 
Hence,  R^x  AD=AO  sin  a  sin  C. 

Equating  this  with  the  value  of  R^  x  AD,  before  found,  and  di- 
viding by  AO,  we  have 

•     -o.       •         •    T»        sin  C     sin  c 
sin  0  sin  C=sin  c  sin  B,  or  - — 1^=- — r     (  1 ) 

sin  B     sinh     ^    ' 

or,         sin  B  :  sin  C  : :  sin  5  :  c        that  is, 

The  sines  of  the  angles  of  a  spherical  triangle  are  to  each 
other  as  the  sines  of  their  opposite  sides, 

IV.  From  K  draw  KE  perpendicular  OB,  and  from  D  draw 
DF  parallel  to  OB.  Then  will  the  angle  DKFr=COB=a, 
since  each  is  the  complement  of  the  angle  EKO. 

In  the  right  angled  triangle  OAH,  we  have 

R  :  cos  c  : :  OA  :  OH  ;  hence 

AO  cos  c=RxOH=RxOE  +  R.DF. 

In  the  right-angled  triangle  OKE 

R  :  cos  a  : :  OK  :  OE,  or  RxOEzrrOK  cos  a. 
But  in  the  right  angled  triangle  OKLA. 

R  :  cos  6  : :  OA  :  OK,  or,  R  x  OK=OA  cos  6. 

TT              w     rwn     i^  A  <^os  a  cos  b 
Hence      RxOE=OA. p 

In  the  right-angled  triangle  KFD 

R  :  sin  a  :  KD  :  DF,  or  R  x  DF=KD  sin  a. 

But  in  the  right  angled  triangles  OAK,  ADK,  we  have 
R  :  sin  5  : :  OA  :  AK,  or  Rx  AK=OA  sin  b 
R  :  cos  K  :  AK  :  KD,  or  RxKD-AK  cos  C 

,                  „^     OA  sin  b  cos  C 
hence         KD= ^ ,  and 

_     _  ^     OA  sin  a  sin  b  cos  C 
RxDF= o2 :  therefore 

-.  .                OA  cos  a  cos  b     AO  sin  a  sin  b  cos  C 
OA  cos  c= ^ + ^2 »  or 

R*  cos  c=R  cos  a  cos  6  + sin  a  sin  b  cos  C. 


248  SPHERICAL  TRIGONOMETRY. 

Similar  equations  may  be  deduced  for  each  of  the  other 
sides.     Hence,  generally, 

R2  cos  a=R  cos  b  cos  c  +  sin  b  sin  c  cos  A.    ) 

R2  cos  b='R  cos  a  cos  c  +  sin  a  sin  c  cos  B.     >      (2.) 

R2  cos  c=R  cos  b  cos  a  +  sin  b  sin  c  cos  B.    ) 

That  is,  radius  square  into  the  cosine  of  either  side  of  a  spheH- 
cal  triangle  is  equal  to  radius  into  the  rectangle  of  the  cosines  of 
the  two  other  sides  plus  the  rectangle  of  the  sines  of  those  sides 
into  the  cosine  of  their  included  angle. 

V.  Each  of  the  formulas  designated  (2)  involves  the  three 
sides  of  the  triangle  together  vv^ith  one  of  the  angles.  These 
formulas  are  used  to  determine  the  angles  when  the  three  sides 
are  known.  It  is  necessary,  however,  to  put  them  under  an- 
other form  to  adapt  them  to  logarithmic  computation. 

Taking  the  first  equation,  we  have 

.      R^  cos  a — R  cos  b  cos  c 

cos  A= : — r-- 

sm  0  sm  c 

Adding  R  to  each  member,  we  have 

R2  cos  a  +  R  sin  b  sin  c — R  cos  b  cos  c 


R  +  cos  A: 


sin  b  sin  c 


But,    R+cos  A^^  cQ^'?A    (^rt.  XXIII.),  and 

R  sin  b  sin  c — R  cos  b  cos  c=— R^  cos  (b  +  c)  (Art,  XIX.) ; 

,  2  cos^^A_R2  (cos  a — cos  (b  +  c)) 

nence,       ■!-»       —  =    t    » z^ 

K  sm  0  sm  c 

^  ^  sini  (u  +  b  +  c)  sini  (b+c-a) 

sm  0  sm  c  ^  ' 

Putting  5 =a+Z>  +  c,    we  shall  have 

|5=i(a+6  +  c)  and  ^s — a=^(b  +  c — a)  ;  hence 

cos  i  j.^^^^HsWS^EK'^ 

^  ^  sm  0  sm  c 


cos  \  B^R y/"^"  ^  (  "  )  ^^".  (^'!=S  ^  (3.) 


sm  a  sm  c 


cos 


1  r  _TJ4.  /^^"  i  (  g  )  sin  (^  s—c) 
^i.-KV  sin  a  sin  6  J 


SPHERICAL  TRIGONOMETRY. 


249 


Had  we  subtracted  each  member  of  the  first  equation  from 
R,  instead  of  adding,  we  should,  by  making  similar  reductions, 
have  found 


sm 


sm 


sin 


_  .  /sin  i(a  4-  b—c)  sin  -^  (a  +  c~b) "^ 


sm  c 


J  15  — K  V  gjjj  ^gjjj  ^ 

^     ~  sin  a  sin  b 


>(4.) 


Putting  5=a4-6  +  c,  we  shall  have 

J« — rt=J(&4-c — a),  ^s — b=h  (a  +  €—b),  and  ^s — c=^{a-^b — c) 

hence, 


sin  1A=R^ /«^"  a^-^)  si"  a^-^) 

sin  6  sin  c 
sin  ^R = R  ^/^nrgg^I^^n"  (j^g^  I  (5.) 


sm  a  sm  c 


sin  y.^R^/s"^  (^^— ^)  si"  (i^— ») 
^  sin  a  sin  b 

VI.  We  may  deduce  the  value  of  the  side  of  a  triangle  in 
terms  of  the  three  angles  by  applying  equations  (4.),  to 
the  polar  triangle.  Thus,  if  «',  b',  c\  A',  B',  C',  represent  the 
sides  and  angles  of  the  polar  triangle,  we  shall  have 

A=180°— a',  B  =  180°— 6',  C=180°— c' ; 
a=l80°—A\  6=180°— B',  and  c=180°— C 

(Book  IX.  Prop.  VII.) :  hence,  omitting  the  ',  since  the  equa- 
tions   are    applicable  to  any  triangle,  we  shall  have 


cos  }.=R^/^^^I^  +  B-C)  '^^^  i  (A  +  C-B)' 


sin  B  sin  C 


cos  i  h=n^Ao^  i  (A  +  B-C)  cos  i  (B  +  C-A) 

sin  A  sin  C 

cos  I  c=Ry/cos  i  (A+C-B)  cos  j  (B+C— A) 

sin  A  sin  B. 
32 


he-) 


250 


SPHERICAL  TRIGONOMETRY. 


Putting  S=A  +  B  +  C,  we  shall  have 

JS~A =i{A  +  B— A),  ^S— B =i  (A  +  C— B)  and 
JS— C=KA+B— C),  hence 


cos 


J^a=R\/'="^  (tS-C)  cos  gS-B) 
sin  B  sin  C 


cos  JL&=R^A°«  (JS-C)  cos  gS-A) 
sin  A  sin  C 

^    /cos(iS-B)cosaS-A) 

COS|C=R  V    : r— 

sm  A  sm  B 


M^O 


VII.  If  we  apply  equations  (2.)  to  the  polar  triangle,  we 
shall  have 

— R^  cos  A'=R  cos  B'  cos  C — sin  B'  sin  C  cos  a\ 

Or,  omitting  the  ',  since  the  equation  is  applicable  to  any  tri- 
angle, we  have  the  three  symmetrical  equations, 


R^.cos  A=sin  B  sin  C  cos  a — ^R  cos  B  cos  C 
R^.cos  B=sin  A  sin  C  cos  b-^R  cos  A  cos  C  W8.) 
R^.cos  C=sin  A  sin  B  cos  c — R  cos  A  cos 


b) 


That  is,  radius  square  into  the  cosine  of  either  angle  of  a  sphe- 
rical triangle^  is  equal  to  the  rectangle  of  the  sines  of  the  two  other 
angles  into  the  cosine  of  their  included  side,  minus  radius  into  the 
rectangle  of  their  cosines. 

VIII.  All  the  formulas  necessary  for  the  solution  of  spheri- 
cal triangles,  may  be  deduced  from  equations  marked  (2.).  If 
we  substitute  for  cos  h  in  the  third  equation,  its  value  taken 
from  the  second,  and  substitute  for  cos^  a  its  value  R^ — sin^  a, 
and  then  divide  by  the  common  factor  R.sin  «,  we  shall  have 

R.cos  c  sin  a:zi^\n  c  cos  a  cos  B  + R.sin  h  cos  C 

T,  .  X-      /,  \    •         .    ,     sin  B  sin  c 

But  equation  (1.)  gives  sm  h= .         >  ; 

sm  C 

hence,  by  substitution, 

R  cos  c  sin  a=sin  c  cos  a  cos  B-f  R. 

Dividing  by  sin  c,  we  have 

cos  c 
R  - —  sin  a=cos  a  cos  B  +  R 
sm  c 


sin  B  cos  C  sin  c 
sin  C 

sin  B  cos  C 


sin  C 


SPHERICAL  TRIGONOMETRY.  251 

Bat,  ^=~  (Art.  XVII.). 
sin      R 

Tlierefore,         cot  c  sin  a=cos  a  cos  B  +  cot  C  «in  B. 

Hence,  we  may  write  the  three  symmetrical  equations, 

cot  a  sin  b=cos  b  cos  C  +  cot  A  sin  C  -x 
cot  b  sin  c=cos  c  cos  A  +  cot  B  sin  A  >  (9.) 
cot  c  sin  azrcos  a  cos  B  +  cot  C  sin  B  / 

That  is,  in  every  spherical  triangle,  the  cotangent  of  one  of  the 
sides  into  the  sine  of  a  second  side,  is  equal  to  the  cosine  of  the  se- 
cond side  into  the  cosine  of  the  included  angle,  plus  the  cotangent 
of  the  angle  opposite  the  first  side  into  the  sine  of  the  included 
angle. 

IX.  We  shall  terminate  these  formulas  by  demonstrating 
Napier^ s  Analogies,  which  serve  to  simplify  several  cases  in  the 
solution  of  spherical  triangles. 

If  from  the  first  equations  (2.)  cos  c  be  eliminated,  there  will 
result,  after  a  little  reduction, 

R  cos  A  sin  c=R  cos  a  sin  b — cos  C  sin  a  cos  b. 

By  a  simple  permutation,  this  gives 

R  cos  B  sin  c=R  cos  b  sin  a — cos  C  sin  b  cos  a. 

Hence  by  adding  these  two  equations,  and  reducing,  we  shall 
have 

sin  c  (cos  A+cos  B)=(R — cos  C)  sin  (a+&) 

^       .  sin   c     sin  «     sin  ft  ^    ^^  . 

But  smce     -: — ri=- — K=- — ^,  we  shall  have 
sm  C     sm  A     sm  B 

sin  c  (sin  A  +  sin  B)  =  sin  C  (sin  a  +  sin  b),    and 

sin  c  (sin  A — sin  B)=siri  C  (sin  a — sin  b),  . 

Dividing  these  two  equations  successively  by  the  preceding 

one  ;  we  shall  have 

sin  A  +  sin  B  _     sin  C         sin  a  +  sin  b 
cosA+cosB~R — cos  C  *  sin  (a+b) 
sin  A — sin  B_     sin  C         sin  a — sin  b 
cos  A+cos  B~R — cos  C  '     sin  (a+bj' 


252  SPHERICAL  TRIGONOMETRY. 

And  reducing  these  by  the  formulas  in  Articles  XXIII.  and 
XXIV.,  there  will  result 

tangi(A+B)=cotJC.5S^ii^ 
°  ^  ^  ^     COS  J(a  +  fe) 

tang  J  (A— B)  =cot  W.-^-Vt r(. 

^^^  ^  ^     sin^(a  +  6) 

Hence,  two  sides  a  and  i  with  the  included  angle  C  being 
given,  the  two  other  angles  A  and  B  may  be  found  by  the 
analogies, 

cosJ(a+6)  :  cos^(a — b)  :  :  cot  |^  C  :  tangJ(A  +  B) 

sin  ^  {a-\-b)  :  sin  J  {a — b)  :  :  cot  ^  C  :  tang  J  (A — B). 

If  these  same  analogies  are  applied  to  the  polar  triangle  of 
ABC,  we  shall  have  to  put  180°— A',  180°— B',  180°— «',  180^— 6', 
180° — c',  instead  of  a,  6,  A,  B,  C,  respectively;  and  for  the  result, 
we  shall  have  after  omitting  the  ',  these  two  analogies, 

cosJ(A  +  B)  :  cosJ(A — B)  :  :  tangle  :  tangJ(rt-{-6) 

sinJ(A  +  B)  :  sin^(A— B)  :  :  tangjc  :  tangj(« — 6), 
by  means  of  which,  when  a  side  c  and  the  two  adjacent  angles 
A  and  B  are  given,  we  are  enabled  to  find  the  two  other  sides 
a  and  b.     These  foiir  proportions  are  known  by  the  name  of 
Napier's  Analogies, 

X.  In  the  case  in  which  there  are  given  two  sides  and  an 
angle  opposite  one  of  them,  there  will  in  general  be  two  solu- 
tions corresponding  to  the  two  results  in  Case  II.  of  rectilineal 
triangles.  It  is  also  plain  that  this  ambiguity  will  extend  itself 
to  the  corresponding  case  of  the  polar  triangle,  that  is,  to  the 
case  in  which  there  are  given  two  angles  and  a  side  opposite 
one  of  them.  In  every  case  we  shall  avoid  all  false  solutions 
by  recollecting, 

1st.  That  every  angle,  and  every  side  of  a  spherical  triangle 
is  less  than  180°. 

2d.  That  the  greater  angle  lies  opposite  the  greater  side,  and 
the  least  angle  opposite  the  least  side,  and  reciprocally. 


NAPIER'S  CIRCULAR  PARTS. 

XI.  Besides  the  analogies  of  Napier  already  demonstrated, 
that  Geometer  also  invented  rules  for  the  solution  of  all  the 
cases  of  right  angled  spherical  triangles. 


SPHERICAL  TRIGONOMETRY. 

In  every  right  angled  spherical 
triangle  BAG,  there  are  six  parts : 
three  sides  and  three  angles.  If 
we  omit  the  consideration  of  the 
right  angle,  which  is  always 
known,  there  will  be  five  remain- 
ing parts,  two  of  which  must 
be  given  before  the  others  can 
be  determined. 

The  circular  parts,  as  they  are  called,  are  the  two  sides  c  and  6, 
about  the  right  angle,  the  complements  of  the  oblique  angles  B 
and  G,  and  the  complement  of  the  hypothenuse  a.  Hence  there 
are  five  circular  parts.  The  right  angle  A  not  being  a  circular 
part,  is  supposed  not  to  separate  the  circular  parts  c  and  &,  so 
that  these  parts  are  considered  as  adjacent  to  each  other. 

If  any  two  parts  of  the  triangle  be  given,  their  corresponding 
circular  parts  will  also  be  known,  and  •  these  together  with  a 
required  part,  will  make  three  parts  under  consideration.  Now, 
these  three  parts  will  all  lie  together,  or  one  of  them  will  be  sepa- 
rated from  both  of  the  others.  For  example,  if  B  and  c  were 
given,  and  a  required,  the  three  parts  considered  would  lie 
together.  But  if  B  and  G  were  given,  and  b  required,  the  parts 
would  not  lie  together ;  for,  B  would  be  separated  from  C  by 
the  part  a,  and  from  b  by  the  part  c.  In  either  case  B  is  the 
middle  part.  Hence,  when  there  are  three  of  the  circular  parts 
under  consideration,  the  middle  part  is  that  one  of  them  to  which 
both  of  the  others  are  adjacent,  or  from  which  both  of  them  are 
separated.  In  the  former  case  the  parts  are  said  to  be  adjacent^ 
and  in  the  latter  case  the  parts  are  said  to  be  opposite. 

This  being  premised,  we  are  now  to  prove  the  following 
rules  for  the  solution  of  right  angled  spherical  triangles,  which 
it  must  be  remembered  apply  to  the  circular  parts,  as  already 
defined. 

1st.  Radius  into  the  sine  of  the  middle  part  is  equal  to  the  rect- 
angle of  the  tangents  of  the  adjacent  parts. 

2d.  Radius  into  the  sine  of  the  middle  part  is  equal  to  the  red- 
angle  of  the  cosines  of  the  opposite  parts. 

These  rules  are  proved  by  assuming  each  of  the  five  circu- 
lar parts,  in  succession,  as  the  middle  part,  and  by  taking  the 
extremes  first  opposite,  then  adjacent.  Having  thus  fixed  the 
three  parts  which  are  to  be  considered,  take  that  one  of  the 
general  equations  for  oblique  angled  triangles,  which  shall  con- 
tain the  three  corresponding  parts  of  the  triangle,  together  with 
the  right  angle  :  then  make  A  =  90°,  and  after  making  the  reduc- 
tions corresponding  to  this  supposition,  the  resulting  equation 
will  prove  the  rule  for  that  particular  case. 


i.  -  y 


'k     254  SPHERICAL  TRIGONOMETRY. 

For  example,  let  comp.  a  be  the  middle  part  and  the  ex- 
tremes opposite.  The  equation  to  be  applied  in  this  case  must 
contain  a,  b,  c,  and  A.  The  first  of  equations  (2.)  contains  these 
four  quantities :  hence 

R2  cos  fl=R  cos  b  cos  c  +  sin  6  sin  c  cos  A. 

If  A = 90°  cos  A=:0 ;  hence 

R  cos  a=cos  b  cos  c ; 

that  is,  radius  into  the  sine  of  the  middle  part,  (which  is  the 
complement  of  a,)  is  equal  to  the  rectangle  of  the  cosines  of  the 
opposite  parts. 

Suppose  now  that  the  complement 
of  a  were  the  middle  part  and  the  ex- 
tremes adjacent.  The  equation  to  be 
appUed  must  contain  the  four  quan- 
tities fl,  B,  C,  and  A.     It  is  the  first 

of  equations  (8.).  

c 

R^  cos  A=sin  B  sin  C  cos  a — R  cos  B  cos  C. 

Making  A =90°,  we  have 

sin  B  sin  C  cos  a=R  cos  B  cos  C,    or 

R  cos  a=cot  B  cot  C  ; 

that  is,  radius  into  the  sine  of  the  middle  part  is  equal  to  the 
rectangle  of  the  tangent  of  the  complement  of  B  into  the  tan- 
gent of  the  complement  of  C,  that  is,  to  the  rectangle  of  the 
tangents  of  the  adjacent  circular  parts. 

Let  us  now  take  the  comp.  B,  for  the  middle  part  and  the 
extremes  opposite.  The  two  other  parts  under  consideration 
will  then  be  the  perpendicular  b  and  the  angle  C.  The  equation 
to  be  applied  must  contain  the  four  parts  A,  B,  C,  and  6 :  it  is  the 
second  of  equations  (8.), 

R^  cos  B=sin  A  sin  C  cos  b — R  cos  A  cos  C. 

Making  A =90°,  we  have,  after  dividing  by  R, 

R  cos  B=sin  C  cos  6. 

Let  comp.  B  be  still  the  middle  part  and  the  extremes  adja- 
cent. The  equation  to  be  applied  must  then  contain  the  four 
four  parts  a,  B,  c,  and  A.     It  is  similar  to  equations  (7.). 

cot  a  sin  c=cos  c  cos  B  +  cot  A  sin  B. 

But  if  A =90°,  cot  A=0 ;  hence, 

cot  a  sin  crrcos  c  cos  B  ;     or, 

R  cos  Brrrcot  a  tang  c. 


SPHERICAL  TRIGONOMETRY.  255 

And  by  pursuing  the  same  method  of  demonstration  when  each 
circular  part  is  made  the  middle  part,  we  obtain  the  five  fol- 
lowing equations,  which  embrace  all  the  cases. 

R  cos  a=:cos  b  cos  c=cot  B  cot  C 
R  cos  Birrcos  b  sin  C==cot  a  tang  c 
R  cos  C  =  cos  csinB=:cot  <2  tang  6 
R  sin  6=sin  <z  sin  B= tang  c  cot  C 
R  sin  c=sin  a  sin  C=:tang6cotBJ 


(10.) 


We  see  from  these  equations  that,  if  the  middle  part  is  required 
we  must  begi7i  the  proportion  with  radius  ;  and  when  one  of  the 
extremes  is  required  we  must  begin  the  proportion  with  the  other 
extreme. 

We  also  conclude,  from  the  first  of  the  equations,  that  when 
the  hy  pothenuse  is  less  than  90^,  the  sides  b  and  c  will  be  of  the  same 
species,  and  also  that  the  angles  B  and  C  will  likewise  be  of  the 
same  species.  When  a  is  greater  than  90°,  the  sides  b  and  c  will 
be  of  different  species,  and  the  same  will  be  true  of  the  angles  B 
and  C.  We  also  see  from  the  two  last  equations  that  a  side  and 
its  opposite  angle  will  always  be  of  the  same  species. 

These  properties  are  proved  by  considering  the  algebraic 
signs  which  have  been  attributed  to  the  trigonometrical  lines, 
and  by  remembering  that  the  two  members  of  an  equation  must 
always  have  the  same  algebraic  sign. 


SOLUTION   OF   RIGHT    ANGLED    SPHERICAL   TRIANGLES    BY 
LOGARITHMS. 

It  is  to  be  observed,  that  when  any  element  is  discovered  in 
the  form  of  its  sine  only,  there  may  be  two  values  for  this  ele- 
ment, and  consequently  two  triangles  that  will  satisfy  the  ques- 
tion ;  because,  the  same  sine  which  corresponds  to  an  angle  or 
an  arc,  corresponds  likewise  to  its  supplement.  This  will  not 
take  place,  when  the  unknown  quantity  is  determined  by  means 
of  its  cosine,  its  tangent,  or  cotangent.  In  all  these  cases,  the 
sign  will  enable  us  to  decide  whether  the  element  in  question  is 
less  or  greater  than  90° ;  the  element  will  be  less  than  90°,  if  its 
cosine,  tangent,  or  cotangent,  has  the  sign  +  ;  it  will  be  greater 
if  one  of  these  quantities  has  the  sign  — . 

In  order  to  discover  the  species  of  the  required  element  of 
the  triangle,  we  shall  annex  the  minus  sign  to  the  logarithms  of 
all  the  elements  whose  cosines,  tangents,  or  cotangents,  are 
negative.     Then  by  recollecting  that  the  product  of  the  two 


256  SPHERICAL  TRIGONOMETRY. 

extremes  has  the  same  sign  as  that  of  the  means,  we  can  at  once 
determine  the  sign  which  is  to  be  given  to  the  required  element, 
and  then  its  species  will  be  known. 

EXAMPLES. 

I.  In  the  right  angled  spherical  tri- 
angle BAG,  right  angled  at  A,  there 
are  given  a =64°  40'  andZ)=42°  12': 
required  the  remaining  parts. 

First,  to  find  the  side  c.  B 

c 

The  hypothenuse  a  corresponds  to  the  middle  part,  and  the 
extremes  are  adjacent :  hence 

R  cos  a=cos  h  cos  c,     or 

As  cos       h  42°  12'         ar.-comp.  log.  0.130296 

Is  to          R  -         -         -         -  -  -  10.000000 

So  is  cos    a  64°  40'         -         -  -  -  9.631326 

Tocos        c  54°  42' 53"            -  -  -  9.761622 

To  find  the  angle  B. 

The  side  h  will  be  the  middle  part  and  the  extremes  oppo- 
site ;  hence 

R  sin  &=cos  (comp.  a)  x  cos  (comp.  B)=sin  a  sin  B. 

As  sin  a  64°  40'  ar.-comp.  log.  0.043911 
Is  to  sin  h  42°  12'  -  -  -  -  9.827189 
Sois       R 10.000000 

To  sin     B       48°  00'  12"  -         -         -         -         9.871100 


To  find  the  angle  C. 

The  angle  C  is  the  middle  part  and  the  extremes  adjacent : 
hence 

B  cos  C=cot  a  tang  &. 

As              R  -  ar.-comp.  log.  0.000000 

Is  to  cot      a  64°  40'  -         -  -  -  9.675237 

So  is  tang  &  42°  12'  -         -  -  -  9.957485 

To  cos        C  64°  34'  -         -  -  -  9.632722 

2.  In  a  right  angled  triangle  BAG,  there  are  given  the  hy- 
pothenuse a=105°  34',  and  the  angle  B=80°  40' :  required  the 
remaining  parts. 


SPHERICAL  TRIGONOMETRY.  257 

To  find  the  angle  C. 

The  hypothenuse  will  be  the  middle  part  and  the  extremes 
adjacent :  hence, 

R  cos  a=cot  B  cot  C. 

ar.-comp.      log.       0.784220  + 
9.428717— 

-  10.0000004- 

-  10.212937— 


As  cot 

B     80°  40' 

Is  to  cos 

a  105°  34'    - 

So  is 

R 

To  cot 

C  148°  30' 54" 

Since  the  cotangent  of  C  is  negative  the  angle  C  is  greater 
than  90°,  and  is  the  supplement  of  the  arc  which  would  corres- 
pond to  the  cotangent,  if  it  were  positive. 

To  find  the  side  c. 
The  angle  B  will  correspond  to  the  middle  part,  and  the 
extremes  will  be  adjacent :  hence, 

R  cos  B=cot  a  tang  c. 

Ascot        a  105°  34'         ar.-comp.         log.  0.555053 — 

Is  to          R              10.000000  + 

So  is  cos  B  80°  40'          -         -         -         .  9.209992  + 

To  tang      c  149°  47' 36"          -         -         -  9.765045^ 

To  find  the  side  h. 

The  side  h  will  be  the  middle  part  and  the  extremes  oppo- 
site: hence, 

Rsin6=sinasin  B. 

As  R       -         ar.  comp.         log.         -  0.000000 

To  sin       a  105°  34'  -         -         -         -  9.983770 

So  is  sin  B     80°  40'  ....  9.994212 

To  sin        h  71°54'  33"       .         .         .         .  9.977982 


OF  QUADRANTAL  TRIANGLES. 

A  quadrantal  spherical  triangle  is  one  which  has  one  of  its 
sides  equal  to  90°. 

Let  BAG  be  a  quadrantal  triangle 
in  which  the  side  «=90°.  If  we  pass 
to  the  corresponding  polar  triangle, 
we  shall  have  A'  =  180°— «=90°,  B'  = 
180°— 6,  G'  =  180°— c,  a'=180°— A, 
^'  =  180°— B,c'  =  180°— G;  from  which 
we  see,  that  the  polar  triangle  will  be 

Y*30 


2^  SPHERICAL  TRIGONOMETRY. 

right  angled  at  A',  and  hence  every  case  may  be  referred  to 
a  right  angled  triangle. 

But  we  can  solve  the  quadrantal  triangle  by  means  of  the 
right  angled  triangle  in  a  manner  still  more  simple. 

In  the  quadrantal  triangle  BAC,  C 

in  which  BC  =  90°,  produce  the  side  >A 

CA  till  CD  is  equal  to  90°,  and  con-  /    \ 

ceive  the  arc  of  a  great  circle  to  be  y^        \ 

drawn  through  B  and  D.    Then  C  ^y^  \ 

will  be  the  pole  of  the  arc  BD,  and         ^^^^.-^  _--J  A 

the  angle  C  will  be  measured  by   B-^I!. .^^ —     \ 

BD  (Book  IX.  Prop.  VI.),  and  the  \....      ^  \b 

angles  CBD  and  D  will  be  right  an-  *'J^*---^ / 

gles.     Now  before  the  remaining  "l) 

parts  of  the  quadrantal  triangle  can 

be  found,  at  least  two  parts  must  be  given  in  addition  to  the 
side  BC  =  90°  ;  in  which  case  two  parts  of  the  right  angled  tri- 
angle BDA,  together  with  the  right  angle,  become  known. 
Hence  the  conditions  which  enable  us  to  determine  one  of  these 
triangles,  will  enable  us  also  to  determine  the  other. 

3.  In  the  quadrantal  triangle  BCA,  there  are  given  CB=r90°, 
the  angle  C=42°  12',  and  the  angle  A=115°  20' :  required' the 
remaining  parts. 

Having  produced  CA  to  D,  making  CD =90°  and  drawn  the 
arc  BD,  there  will  then  be  given  in  the  right  angled  triangle 
BAD,  the  side  fl=C=42°  12',  and  the  angle  BAD=180°— 
BAC  =  180°— 115°  20'  =  64°40',to  find  the  remaining  parts. 

To  find  the  side  d. 

The  side  a  will  be  the  middle  part,  and  the  extremes  oppo« 
site:  hence, 

R  sin  <2  =  sin  A  sin  d. 

As  sin     A  64°  40'               ar.-comp.       log.  0.043911 

Isto        R 10.000000 

So  is  sin  a  42°  12'          .         -         -         -  9.827189 

To  sin      d  48°  00' 15"    -                 -         -  9.871100 

To  find  the  angle  B. 

The  angle  A  will  correspond  to  the  middle  part,  and  the  ex- 
tremes will  be  opposite  :  hence 

R  cos  A=sin  B  cos  «. 

As  cos       a     42°  12'  ar.-comp.       log.         0.130296 

Isto         R       -         -         -        -         -         -       10.000000 

So  is  cos  A     64°  40'  ....         9.631326 

To  sin       B     35°  16' 53"  -        -        -         9^761622 


SPHERICAL  TRIGONOMETRY.  269 

To  find  the  side  b. 

The  side  b  will  be  the  middle  part,  and  the  extremes  adja- 
cent :  hence, 

R  sin  b=coi  A  tang  a. 

As             R             -  ar.-comp.  log.  0.000000 

Is  to  cot  A  64°  40'  -         -         .  -  9.675237 

So  is  tang  a  42°  12'  -         -         -  -  9.957485 

To  sin        b  25°  25'  14"             -         -  -  9.632722 

Hence,     CA=90°— 6=90°— 25°  25'  14"  =64°  34'  46" 

CBAr=90°— ABD=90°— 35°  16'  53 '=54°  43  07" 
BA^d  -         -         .         -         =48°  00' 15". 

4.  In  the  right  angled  triangle  BAC,  right  angled  at  A,  there 
are  given  a=115°  25',  and  c=60°  59'  i^-equired  the  remaining 
parts. 

(  B=148°  56'  45" 

Ans.    )  C=  75°  30'  33" 

(  6=152°  13' 50". 

6.  In  the  right  angled  spherical  triangle  BAC,  right  angled 
at  A,  there  are  given  c=116°  30'  43",  and  6=29°  41'  32" :  re- 
quired the  remaining  parts. 

(  C=:103°  52'  46" 

Ans,    )  B=  57°  28'  22" 

(a  =112°  47' 58". 

6.  In  a  quadrantal  triangle,  there  are  given  the  quadrantal 

side  ==90°,  an  adjacent  side  =115°  09',  and  the  included  angle 

=  115°  55' ;  required  the  remaining  parts. 

side,  113°  18' 19" 

^w*-    ^  o^^io.    S  117°  33' 52" 
angles,      ^^j^  4Q.  q^.. 


SOLUTION  OF  OBLIQUE  ANGLED  TRIANGLES  BY  LOGARITHMS. 

There  are  six  cases  which  occur  in  the  solution  of  oblique 
angled  spherical  triangles. 

1.  Having  given  two  sides,  and  an  angle  opposite  one  of 
them. 

2.  Having  given  two  angles,  and    a  side  opposite  one  of 
them. 

3.  Having  given  the  three  sides  of  a  triangle,  to  find  the 
angles. 


260  SPHERICAL  TRIGONOMETRY. 

4.  Having  given  the  three  angles  of  a  triangle,  to  find  the 
sides. 

5.  Having  given  two  sides  and  the  included  angle. 

6.  Having  given  two  angles  and  the  included  side. 


CASE  I. 

Cfiven  two  sides ,  and  an  angle  opposite  one  of  them,  to  find  the  re- 
maining parts. 

For  this  case  we  employ  equation  {!.)  ; 

As  sin  a  :  sin  6  :  :  sin  A  :  sin  B. 

• 

Ex.  1.  Given  the  side  a=44° 
13' 45",  &z=84°  14'  29"  and  the 
angle  A=32°  26' 07"  :  required 
the  remaining  parts. 

To  find  the  angle  B. 

As  sin       a  44°  13' 45"      ar.-comp.         log.  0.156427 

Is  to  sin    b  84°  14'  29"  -         -         -  9.997803 

So  is  sin  A  32°  26'  07"  -         -         -  9.729445 

To  sin      B  49°  54'  38"  or  sin  B'  130°  5'  22"  9.883685 

Since  the  sine  of  an  arc  is  the  same  as  the  sine  of  its  supple- 
ment, there  will  be  two  angles  corresponding  to  the  logarithmic 
sine  9.883685  and  these  angles  will  be  supplements  of  each 
other.  It  does  not  follow  however  that  both  of  them  will  satisfy 
all  the  other  conditions  of  the  question.  If  they  do,  there  will 
be  two  triangles  ACB',  ACB  ;  if  not,  there  will  be  but  one. 

To  determine  the  circumstances  under  which  this  ambiguity 
arises,  we  will  consider  the  2d  of  equations  (2.). 

R^  cos  b=R  cos  a  cos  c+sin  a  sin  c  cos  B. 

from  which  we  obtain 

_,     R^  cos  b — R  cos  0!  cos  c 

cos  B  = : -. . 

sm  a  sm  c 

Now  if  cos  b  be  greater  than  cos  a  or  cos  c,  we  shall  have 

R^  cos  6>R  cos  a  cos  c, 

or  the  sign  of  the  second  member  of  the  equation  will  depend 
on  that  of  cos  b.    Hence  cos  B  and  cos  b  will  have  the  same 


SPHERICAL  TRIGONOMETRY.  261 

sign,  or  B  and  b  will  be  of  the  same  species,  and  there  will  be 
but  one  triangle. 

But  when     cos  fe>cos  a,  sin  ft  <  sin  a  :  hence, 

If  the  sine  of  the  side  opposite  the  required  angle  be  less  than 
the  sine  of  the  other  given  side,  there  will  be  but  one  triangle. 

If  however,  sin  ft  >  sin  a,  the  cos  ft  will  be  less  than  cos  a, 
and  it  is  plain  that  such  a  value  may  then  be  given  to  c  as  to 
render 

R^cos  ft<R  cos  a  cos  c, 

or  the  sign  of  the  second  member  may  be  made  to  depend  on 
cos  c. 

We  can  therefore  give  such  values  to  c  as  to  satisfy  the  two 
equations 

_     R^  cos  ft — R  cos  a  cos  c 
-l-cos  B=— 


— cos  B: 


sm  a  sm  c 

R^  cos  ft — R  cos  a  cos  c 

sin  a  sin  c 


Hence,  if  the  sine  of  the  side  opposite  the  required  angle  be 
greater  than  the  sine  of  the  other  given  side,  there  will  be  two  tri- 
angles which  will  fulfil  the  given  conditions.' 

Let  us,  however,  consider  the  triangle  ACB,  in  which  we  are 
yet  to  find  the  base  AB  and  the  angle  C.  We  can  find  these 
parts  most  readily  by  dividing  the  triangle  into  two  right  angled 
triangles.  Draw  the  arc  CD  perpendicular  to  the  base  AB  : 
then  in  each  of  the  triangles  there  will  be  given  the  hypothe- 
nuse  and  the  angle  at  the  base.  And  generally,  when  it  is 
proposed  to  solve  an  oblique  angled  triangle  by  means  of  the 
right  angled  triangle,  we  must  so  draw  the  perpendicular  that 
it  shall  pass  through  the  extremity  of  a  given  side,  and  lie  oppo- 
site  to  a  given  angle. 

To  find  the  angle  C,  in  the  triangle  ACD. 

As  cot          A  320  26'  07"     j^r.-comp.       log.  9.803105 

Is  to             R 10.000000 

So  is  cos       ft  84°  14'  29"          -         -         -  9.001465 

To  cot    ACD  86°  21'  09"          -         -         -  8.804570 

To  find  the  angle  C  in  the  triangle  DCB. 

Ascot           B  49°  54' 38"     ar.-comp.     log.  0.074810 

Is  to              R 10.000000 

So  is  cos       a  44°  13'  45"      -         -         -  9.855250 

To  cot     DCB  49°  35' 38"      -         -         -  9.93006Q 

Hence  ACB  =  1 35°  56'  47". 


262  SPHERICAL  TRIGONOMETRY. 


To  find  the  side  AB. 

As  sin       A     32°  26'  07" 
Is  to  sin    C  135°  56' 47" 
So  is  sin     a     44°  13' 45" 

ar.-comp. 

log. 

0.270555 
9.842191 
9.843563 

To  sin         c  115°  16' 29" 

- 

- 

9.956309 

The  arc  64°  43'  31",  which  corresponds  to  sin  c  is  not  the 
value  of  the  side  AB  :  for  the  side  AB  must  be  greater  than  &, 
since  it  lies  opposite  to  a  greater  angle.  But  6=84°  14'  29"  : 
hence  the  side  AB  must  be  the  supplement  of  64°  43'  31",  or 
115°  16' 29". 

Ex.  2.  Given  6=91°  03'  25",  a=40°  36'  37",  and  A=35°  57' 
15" :  required  the  remaining  parts,  when  the  obtuse  angle  B  is 
taken. 

(B  =  115°35'4r' 

^715.    )  C=   58°  30' 57" 

)  c  =  70°  58'  52" 


CASE  n. 

Having  given  two  angles  and  a  side  opposite  one  of  them,  to  find 
the  remaining  parts. 

For  this  case,  we  employ  the  equation  (1.) 
sin  A  :  sin  B  :  :  sin  a  :  sin  b. 

Ex.  1.  In  a  spherical  triangle  ABC,  there  are  given  the  angle 
A =50°  12',  B  =  58°  8',  and  the  side  a =62°  42' ;  to  find  the  re- 
maining parts. 

To  find  the  side  h. 

As  sin       A     50°  12'         ar.-comp.         log..      0.114478 
Is  to  sin    B     58°  08'  -         -         -         -         9.929050 

So  is  sin    a     62°  42'  -         -         .         -         9.948715 


^  To  sin        b     79°  12'  10",  or  100°  47'  50"  9.992243 

We  see  here,  as  in  the  last  example,  that  there  are  two  arcs 
corresponding  to  the  4th  term  of  the  proportion,  and  these  arcs 
are  supplements  of  each  other,  since  they  have  the  same  sine. 
It  does  not  follow,  however,  that  both  of  them  will  satify  all 
the  conditions  of  the  question.  If  they  do,  there  will  be  two 
triangles ;  if  not,  there  will  be  but  one. 


SPHERICAL  TRIGONOMETRY.  203 

To  determine  when  there  are  two  triangles,  and  also  when 
there  is  but  one,  let  us  consider  the  second  of  equations  (8.) 

R*  cos  B=sin  A  sin  C  cos  b — R  cos  A  cos  C,      which  gives 

_     R^  cos  B — R  cos  A  cos  C 

cos  0= = — I — = — r5 . 

sin  A  sin  C 

Now,  if  cos  B  be  greater  than  cos  A  or  cos  C, 

R^  cos  B>R  cos  A  cos  C, 

and  hence  the  sign  of  the  second  member  of  the  equation  will 
depend  on  that  of  cos  B,  and  consequently  cos  b  and  cos  B  will 
have  the  same  algebraic  sign,  or  b  and  B  will  be  of  the  same 
species.     But  when  cos  B  >cos  A  the  sin  B<sin  A  :  hence 

If  the  sine  of  the  angle  opposite  the  required  side  be  less  than 
the  sine  of  the  other  given  angle,  there  will  be  but  one  solution. 

If,  however,  sin  B>sin  A,  the  cos  B  will  be  less  than  cos  A, 
and  it  is  plain  that  such  a  value  may  then  be  given  to  cos  C,  as 
to  render 

R^cos  B<R  cos  A  cos  C, 

or  the  sign  of  the  second  member  of  the  equation  may  be  made 
to  depend  on  cos  C.  We  can  therefore  give  such  values  to  C 
as  to  satisfy  the  two  equations 

_     R^  cos  B — R  cos  A  cos  C 
+  C0S  0= : — 7 — : — j^ ,     and 

sm  A  sm  C 

_     R2  cofe  B — R  cos  A  cos  C 

cos  0= : 7 : pq . 

sm  A  sin  L> 

Hence,  if  the  sine  of  the  angle  opposite  the  required  side  be 
greater  than  the  sine  of  the  other  given  angle  there  will  be  two 
solutions. 

Let  us  first  suppose  the  side  b  to  be  less  than  90°,  or  equal 
to  79°  12'  10". 

If  now,  we  let  fall  from  the  angle  C  a  perpendicular  on  the 
base  BA,  the  triangle  will  be  divided  into  two  right  angled  tri- 
angles, in  each  of  which  there  will  be  two  parts  known  besides 
the  right  angle. 

Calculating  the  parts  by  Napier's  rules  we  find, 

Angle  C=130°54'26" 

Side      c=119°03'26". 

If  we  take  the  side  6=100°  47'  50",  we  shall  find 

Angle  C=156°  15'  04" 

Side      c=:152°  14'  18". 


264  SPHERICAL  TRIGONOMETRY. 

Ex,  2.  In  a  spherical  triangle  ABC  there  are  given  A=103'' 
59'  57",  B=46°  18'  7",  and  a=42°  8'  48"  ;  required  the  remain- 
ing parts. 

There  will  but  one  triangle,  since  sin  B<sin  A. 

/  b  =30° 
Ans.    )  C=36°  7'  53" 
/  c  =24°  3'  55". 


CASE  III. 

Having  given  the  three  sides  of  a  spherical  trangle  to  find  the 

angles. 


For  this  case  we  use  equations  (3.). 


/sin  ^s  sin  {\s—a) 
cosJA=RV  sin  ^  sine 

Ex,  1.  In  an  oblique  angled  spherical  triangle  there  are 
given  a=56°  40',  6=83°  13'  and  c=114°  30';  required  the 
angles. 

J(a  +  6  +  c)=Js        =127°  11'  30" 

J(6  +  c— «)=(J5— a)=70°31'  30". 

Log  sin      Js  127°  11' 30"           -         -         -  9.901250 

log  sin  (Js— a)  70°  31'  30"           -         -         -  9.974413 

—log  sin     h     83°  13'                ar.-comp.  0.003051 

—log  sin     c  114°  30'                ar.-comp.  0.040977 

Sum "l9.'91969l 

Half  sum  =  log  cos  JA  24°  15',  39"     -         -  9.959845 

Hence,  angle  A =48°  31'  18". 

The  addition  of  twice  the  logarithm  of  radius,  or  20,  to  the 
numerator  of  the  quantity  under  the  radical  just  cancels  the  20 
which  is  to  be  subtracted  on  account  of  the  arithmetical  com- 
plements, to  that  the  20,  in  both  cases,  may  be  omitted. 

Applying  the  ^ame  formulas  to  the  angles  B  and  C,  we  find, 

Bnz  62°  55'  46" 
C  =  125°  19' 02". 
Ex,  2.  In  a  spherical  triangle  there  are  given  a:nr40°  18'  29", 
6=67°  14'  28",  and  c=89°  47'  6"  :  required  the  three  angles. 

A=  34°  22'  18" 

Ans,    \  B=  53°  35'  16" 

C  =  119o  13' 32". 


SPHERICAL  TRIGONOMETRY. 


265 


CASE  IV. 

Saving  given  the  three  angles  of  a  spherical  triangle,  to  find  the 
three  sides. 


For  this  case  we  employ  equations  (7.) 

/cosaS-B)cos(^S-C) 

cosJa=RV  .    p    .    ^ . 

sin  Jd  sin  C 

Ex.  L  In  a  spherical  triangle  ABC  there  are  given  A =48° 
30',  B=125°  20',  and  C=62°  54' ;  required  the  sides. 

|(A+B  +  C)=^S=  118°  22' 
=     69°  52' 


(^S-B) 
(^S-C) 


=— .  6°  58' 
=     55°  28' 


Log  cos  {^S— B)  —6°  58' 
log  cos  (|S— C)  55°  28' 
—log  sin  B  125°  20' 
— ^log  sin      C  62°  54' 

Sum 

Half  sum=log  cos  ^A=28°  19'  48" 

Hence,  side  a =56°  39'  36". 

In  a  similar  manner  we  find, 


ar.-comp. 
ar.-comp. 


9.996782 
9.753495 
0.088415 
0.050506 
19.889198 
9.944599 


t= 1140  29'  58" 
c=  83°  12'  06". 


Ex»  2.  In  a  spherical  triangle  ABC,  there  are  given  A =109° 
55'  42",  B  =  116o  38'  33",  and  C=120o  43'  37";  required  the 
tlu'ee  sides. 


Ans, 


a=  98°  21' 40^ 
6  =  109°  50' 22" 
c  =  115°  13' 28". 


CASE  V. 


Having  given  in  a  spherical  triangle,  two  sides  and  their  in- 
cluded angle,  to  find  the  remaining  parts. 


266 


SPHERICAL  TRIGONOMETRY. 


For  this  case  we  employ  the  two  first  of  Napier's  Analogies, 
cos  ^{a-\-b)  :  cos  i(« — b)  :  :  cot  ^C  :  tang  ^(A  +  B) 
sin  ^{a  +  b)  :  sin  ^(a — b)  :  :  cot  ^C  :  tang  i(A — B). 

Having  found  the  half  sum  and  the  half  difference  of  the 
angles  A  and  B,  the  angles  themselves  become  known  ;  for,  the 
greater  angle  is  equal  to  the  half  sum  plus  the  half  difference, 
and  the  lesser  is  equal  to  the  half  sum  minus  the  half  diffe- 
rence. 

The  greater  angle  is  then  to  be  placed  opposite  the  greater 
side.  The  remaining  side  of  the  triangle  can  then  be  found  by 
Case  11. 

Ex.  1.  In  a  spherical  triangle  ABC,  there  are  given  a=68^ 
46'  2",  6=37°  10',  and  c=39°  23' ;  to  find  the  remaining  parts. 

|(«  +  &)  =  52°  58'  1",  J(a— 6)  =  15o  48'  1",  ^0  =  19°  41'  30". 

As  cos      i(a  +  6)52°58'    1"  log.  ar.-comp.  0.220210 

Is  to  cos   i{a—b)  15"  48'    1"      -  .         _  9.983271 

So  is  cot         ^C    19°  41' 30"      -  -         -  10.446254 

Totangi(A  +  B)    77°  22' 25"      -  -         -  10.649735 

As  sin      ^(a  +  b)  52°  58'    1"  log.  ar.-comp.  0.097840 

Is  to  sin  -h{a—b)  15°  48'    1"       -  -         -  9.435016 

So  is  cot      iC      19°  41' 30"  -  -         -  10.446254 

To  tang  i(A— B)  43°  37'  21"  -  -         -  9.979110 

.  Hence,  A=77°  22'  25"  +  43°  37'  21"=120°  59'  46" 
B=77°  22'  25 " — 430  37'  21"=  33°  45'  04" 
side  c         -         -         -         -  =  430  37'  37". 

Ex.  2.  In  a  spherical  triangle  ABC,  there  are  given  b=8S° 
19'  42",  c=23°  27'  46",  the  contained  angle  A=20°  39'  48" ; 
to  find  the  remaining  parts. 

B  =  156°30'17" 
Ans.    )  C=     9°  11^47 
61°  32'  16". 


(B  =  : 


CASE  VI. 


In  a  spherical  triangle,  having  given  two  angles  and  the  included 
side  to  find  the  remaining  parts. 


tf' 


SPHERICAL  TRIGONOMETRY.  267 

For  this  case  we  employ  the  second  of  Napier's  Analogies, 
cos  J(A+B)  :  cos|(A— B)  :  :  tang  Jc  :  tangJ(a+6) 
sinJ(A+B)  :  sin  J  (A— B)  :  :  tang  Jc  ;  tang  J(a— &). 
From  which  a  an^  b  are  found  as  in  the  last  case.    The  re- 
maining angle  can  then  be  found  by  Case  L 

Ex.  1.  In  a  spherical  triangle  ABC,  there  are  given  A=81° 
SS'  20",  B=70''  9'  38",  c=59°  16'  23"  ;  to  find  the  remaining 
parts. 

J(A  +  B)  =75°  53'  59",  J(A— B)  =5°  44'  21",  }c=29°  38'  11". 

As  cos       i(A  +  B)  75°  53' 59"  log.    ar.-comp.  0.613287 

Tocos        KA— B)     5°  44' 21"  -         -         9.997818 

So  is  tang       ic         29°  38' 11"  -         -         9.755051 

To  tang       i(a  +  b)  66°  42'  52"  -         -       10.366156 

As  sin         i(A+B)  75°  53'  59"  log.  ar.-comp.  0.013286 
To  sin        |(A— B)     5°  14'  21"  -         -         9.000000 

So  is  tang       ic         29°  38'  11"  -         -         9.755051 

To  tang       i{a—b)     3°  21' 25"  -         -         8.768337 

Hence       a=66o  42'  52" +  3°  21'  25"=70°  04'  17" 

&=66°  42'  52"-— 3°  21'  25"=63°  21'  27" 

angle  C        -        -        -     =64°  46' 33". 

Ex,  2.  In  a  spherical  triangle  ABC,  there  are  given  A =34° 
15'  3",  B=42°  15'  13",  and  c=76°  35'  36" ;  to  find  the  remain- 
ing parts. 

(  a  =40°    0'  10" 

Ans.    )  5  ==50°  10'  30" 

(  C  =58°  23' 41". 


A   TABLE 


OP 


liOGARITHMS    OF    LUMBERS 

FROM    1    TO    10,000. 


N. 
1 

Log. 
0.000000 

N. 
26 

Lo[r. 

N. 
51 

Log. 

N. 
76 

Log. 
1.880814 

1.414973 

1.707570 

2 

0.301030 

27 

1.431364 

52 

1.716003 

77 

1.886491 

3 

0.477121 

28 

1.447158 

53 

1.724276 

78 

1.892095 

4 

0.602060 

29 

1.462398 

54 

1.732394 

79 

1.897627 

5 

0.698970 

30 

1.477121 

55 

1.740363 

80 

1.903090 

6 

0.778151 

31 

1.491362 

56 

1.748188 

81 

1.908485 

7 

0.845098 

32 

1.505150 

57 

1.755875 

82 

1.913814 

8 

0.903090 

33 

1.518514 

58 

1.763428 

83 

1.919078 

9 

0.954243 

34 

1.531479 

69 

1.770852 

84 

1.924279 

10 

1.000000 

35 

1.544068 

60 

1.778151 

85 

1.929419 

11 

1.041393 

36 

1.556303 

61 

1.785330 

86 

1.934498 

12 

1.079181 

37 

1.568202 

62 

1.792392 

87 

1.939519 

13 

1.113943 

38 

1.579784 

63 

1.799341 

88 

1.944483 

14 

1.146128 

39 

1.591065 

64 

1.806180 

89 

1.949390 

15 

1,176091 

40 

1.602060 

65 

1.812913 

90 

1.954243- 

16 

1.204120 

41 

1.612784 

66 

1.819544 

91 

1.959041 

17 

1.230449 

42 

1.623249 

67 

1.826075 

92 

1.963788 

18 

1.255273 

43 

1.633468 

68 

1.832509 

93 

1.968483 

19 

1.278754 

44 

1.643453 

69 

1.838849 

94 

1.973128 

20 

1.301030 

45 

1.653213 

70 

1.845098 

95 

1.977724 

21 

1.322219 

46 

1.662758 

71 

1.851258 

96 

1.982271 

22 

1.342423 

47 

1.672098 

72 

1.857333 

97 

1.986772 

23 

1.361728 

48 

1.681241 

73 

1.863323 

98 

1.991226 

24 

1.380211 

49 

1.690196 

74 

1.869232 

99 

1.995635 

25 

1.397940 

50 

1.698970 

75 

1.875061 

100 

2.000000 

N.B.  In  the  following  table,  in  the  last  nine  columns  of  each 
page,  where  the  first  or  leading  figures  change  from  9's  to  O's, 
points  or  dots  are  introduced  instead  of  the  O's  through  the  rest 
of  the  line,  to  catch  the  eye,  and  to  indicate  that  from  thence 
the  annexed  first  two  figures  of  the  Logarithm  in  the  second 
column  stand  in  the  next  lower  line. 


2 

A  TABLE  OF  LOGARITHMS  FROM  1 

TO  10,000 

IN.  1   0   1  1  1  2  1  3  i  4  i  5  i  6  1  7  1  8  1  9  1  D.  1 

lioo 

OOOOOOi  0434 

0868 

1301  1734 

2J66 

2598  3029 ! 

3461 

3891 

432 

101 

4321 

4751 

5181 

5609 

6038 

6466 

6^94 

7321 

7748 

8174 

428 

103 

8000 

9020 

9451 

9876 

.300 

.724 

1147 

1570 

1993 

2415 

424 

103 

012837 

3259 

3680 

4100 

4521 

4940 

5360 

5779 

6197 

6616 

419 

104 

7033 

7451 

7868 

8284 

8700 

9116 

9532 

9947 

.361 

.775 

416 

105 

021189 

1603 

2016 

2428 

2841 

3252 

3664 

4075 

4486 

4896 

412 

106 

5306 

5715 

6125 

6533 

6942 

7350 

7757 

8164 

8.571 

8978 

408 

107 

9384 

9789 

.195 

.600 

1004 

1408 

1812 

2216 

2619 

3021 

404 

108 

033424 

3826 

4227 

4628 

5029 

5430 

5830 

6230 

6629 

7028 

400 

109 
110 

7426 

7825 
1787 

8223 
2182 

8620 
2576 

9017 
2969 

9414 
3362 

9811 
3755 

.207 

.602 

.998 
4932 

396 
393 

041393 

4148 

4540 

111 

5323 

5714 

6105 

6495 

6885 

7275 

7664 

8053 

8442 

8830 

389 

112 

9218 

9606 

9993 

.380 

.766 

1153 

1538 

1924 

2309 

2694 

386 

113 

053078 

3463 

3846 

4230 

4613 

4996 

5378 

5760 

6142 

6524 

382 

114 

6905 

7286 

7666 

8046 

8426 

8805 

9185 

9563 

9942 

.320 

379 

115 

060698 

1075 

1452 

1829 

2206 

2582 

2958 

3333 

3709 

4083 

376 

116 

4458 

4832 

5206 

5580 

5953 

6326 

6699 

7071 

7443 

7815 

372 

117 

8186 

8557 

8928 

9298 

9668 

..38 

.407 

.776 

1145 

1514 

369 

118 

071882 

2250 

2617 

2985 

3352 

3718 

40S5 

4451 

4816 

5182 

366 

119 
120 

5547 

5912 
9543 

6276 
9904 

6640 
.286 

7004 
.626 

7368 
.987 

7731 
1347 

8094 
1707 

8457 
2067 

8819 
2426 

363 
360 

079181 

121 

082785 

3144 

3503 

3861 

4219 

4576 

4934 

.5291 

5647 

6004 

357 

122 

6360 

6716 

7071 

7426 

7781 

8136 

8490 

8845 

9198 

9552 

355 

123 

9905 

.258 

.611 

.963 

1315 

1667 

2018 

2370 

2721 

3071 

.351 

124 

093422 

3772 

4122 

4471 

4820 

5169 

.5518 

5866 

6215 

6562 

349 

125 

6910 

7257 

7604 

7951 

8298 

8644 

8990 

9335 

9681 

..26 

346 

126 

100371 

0715 

1059 

1403 

1747 

2091 

2434 

2777 

3119 

3462 

,343 

127 

3804 

4146 

4487 

4828 

5169 

5510 

5851 

6191 

6531 

6871 

340 

128 

7210 

7549 

7888 

8227 

8565 

8903 

9241 

9579 

9916 

.253 

338 

129 
130 

110590 

0926 
4277 

1263 
4611 

1599 
4944 

1934 

5278 

2270 
5611 

Si605 
5943 

2940 
6276 

3275 
6608 

3609 
6940 

335 
333 

113943 

131 

7271 

7603 

7934 

8265 

8595 

8926 

9256 

9586 

9915 

.245 

.330 

132 

120574 

0903 

1231 

1560 

1888 

2216 

2544 

2871 

3198 

3525 

328 

133 

3852 

4178 

4504 

4830 

5156 

5481 

5806 

6131 

6456 

6781 

325 

134 

7105 

7429 

7753 

8076 

8399 

8722 

9045 

9368 

9690 

..12 

.323 

135 

130334 

0655 

0977 

1298 

1619 

1939 

2260 

2580 

2900 

3219 

.321 

136 

3539 

3858 

4177 

4496 

4814 

5133 

5451 

5769 

6086 

6403 

318 

137 

6721 

7037 

7354 

7671 

7987 

8303 

8618 

8934 

9249 

9564 

315 

138 

9879 

.194 

.508 

.822 

1136 

1450 

1763 

2076 

2389 

2702 

314 

139 
140 

143015 

3327 
6438 

3639 
6748 

3951 
7058 

4263 
7367 

4574 
7676 

4885 
7985 

5196 
82M 

.5507 
8603 

5818 
8911 

311 
309 

146128 

141 

9219 

9527 

9835 

.142 

.449 

.756 

1063 

1370 

1676 

1982 

.307 

142 

152288 

2594 

2900 

3205 

3510 

3815 

4120 

4424 

4728 

5032 

30.5^ 

143 

5336 

5840 

5943 

6246 

6549 

6852 

7154 

7457 

7759 

8061 

303 

144 

8362 

8664 

8965 

9266 

9567 

9868 

.168 

.469 

.769 

1068 

301 

145 

161368 

1667 

1967 

2266 

2564 

2863 

3161 

3460 

3758 

4055 

299 

146 

4353 

4650 

4947 

5244 

5541 

5838 

6134 

6430 

6726 

7022 

297 

147 

7317 

7613 

7908 

8203 

8497 

8792 

9086 

9380 

9674 

9968 

295 

148 

170262 

0555 

084S 

1141 

1434 

1726 

2019 

2311 

2603 

2895 

293 

149 
150 

3186 

3478 
6381 

3769 
6670 

4060 
6959 

4351 

7248 

4641 

4932 

5222 
8113 

6512 
8401 

5802 
8689 

291 

289 

176091 

7536 j  7825 

151 

8977 

9264 

9552 

9839 

.126 

.413 

.699 

.985 

1272 

1.558 

287 

152 

181844 

2129 

2415 

2700 

2985 

3270 

3555 

3839 

4123 

4407 

285 

153 

4691 

4975 

5259 

5542 

5825 

6108 

6391 

6674 

6956 

7239 

283 

154 

7521 

7803 

8084 

8366 

8647 

8928 

9209 

9490 

9771 

..51 

281 

155 

190332 

0612 

0892 

1171 

1451 

1730 

2010 

2289 

2567 

2846 

279 

156 

3125 

3403 

3681 

3959 

4237 

4514 

4792 

5069 

5346 

5623 

278 

157 

5899 

6176 

6453  6729 

7005 

7281 

7556 

7832 

8107 

8382 

276 

158 

8657 

8932 

9206  9481 

9755 

..29 

.303 

.577 

.8.50 

1124 

274 

159  '2013971  16701  194312216 

2488 

2761 

3033 

33051  35771  3848  i  272  | 

N.  1   0   |l|2l3|4i5|6|7|8|9|D.  1 

A  TABLE  OP  LOGARITHMS  FROM  1 

ro  10,000. 

3 

N,  !   0   |ll2l3|4l5l6|7l8|9|D.  1 

1»»0 

204120 

4391 

4663 

4934 

5204 

5475 

5746 

6016 

6286| 6556 

271 

161 

6826 

7096 

7365 

7634 

7904 

8173 

8441 

8710 

8979 

9247 

269 

162 

9515 

9783 

..51 

.319 

.586 

.853 

1121 

1388 

1654 

1921 

267 

163 

212188 

2454 

2720 

2986 

3252 

3518 

3783 

4049 

4314 

4579 

266 

164 

4844 

5109 

5373 

6638 

5902 

6166 

6430 

6694 

6957 

7221 

264 

165 

7484 

7747 

8010 

8273 

8536 

8798 

9060 

9323 

9585 

9846 

262 

166 

220108 

0370 

0631 

0892 

1153 

1414 

1675 

1936 

2196 

2456 

261 

167 

2716 

2976 

3236 

3496 

3755 

4015 

4274 

4533 

4792 

5051 

259 

168 

6309 

6568 

5826 

6084 

6342 

6600 

6858 

7115 

7372 

7630 

258 

169 
170 

7887 

8144 

8400 
0960 

8657 
1215 

8913 

9170 
1724 

9426 
1979 

9682 
2234 

9938 

.193 

256 
254 

230449 

0704 

1470 

2488 

2742 

171 

2996 

3250 

3504 

3757 

4011 

4264 

4517 

4770 

5023 

5276 

253 

172 

5528 

5781 

6033 

6285 

6537 

6789 

7041 

7292 

7544 

7795 

252 

173 

8046 

8297 

8548 

8799 

9049 

9299 

9550 

9800 

..50 

.300 

250 

174 

240549 

0799 

1048 

1297 

1546 

1795 

2044 

2293 

2541 

2790 

249 

175 

3038 

3286 

3534 

3782 

4030 

4277 

4525 

4772 

5019 

5266 

248 

176 

5513 

5759 

6006 

6252 

6499 

6745 

6991 

7237 

7482 

7728 

246 

177 

7973 

8219 

8464 

8709 

8954 

9198 

9443 

9687 

9932 

.176 

245 

178 

250420 

0664 

0908 

1151 

1395 

1638 

1881 

2125 

2368 

2610 

243 

179 
180 

2853 

3096 
5514 

3338 
5755 

3580 

3822 
6237 

4064 
6477 

4306 
6718 

4548 
6958 

4790 
7198 

5031 
7439 

242 
241 

255273 

5996 

181 

7679 

7918 

8158 

8398 

8637 

8877 

9116 

9355 

9594 

9833 

239 

182 

260071 

0310 

0548 

0787 

1025 

1263 

1501 

1739 

1976 

2214 

238 

183 

2451 

2688 

2925 

3162 

3399 

3636 

3873 

4109 

4346 

4582 

237 

184 

4818 

5054 

5290 

5525 

5761 

5996 

6232 

6467 

6702 

6937 

235 

185 

7172 

7406 

7641 

7875 

8110 

8344 

8578 

8812 

9046 

9279 

234 

186 

9513 

9746 

9980 

.213 

.446 

.679 

.912 

1144 

1377 

1609 

233 

187 

271842 

2074 

2306 

2538 

2770 

3001 

3233 

3464 

3696 

3927 

232 

188 

4158 

4389 

4620 

4850 

5081 

5311 

5542 

5772 

6002 

6232 

230 

189 
190 

6462 

6692 
8982 

6921 

7151 
9439 

7380 
9667 

7609 

7838 
.123 

8067 
..361 

8296 
.578 

8525 
.806 

229 

228 

278754 

9211 

9895 

191 

281033 

1261 

1488 

1715 

1942 

2169 

2396 

2622 

2849 

3075 

227 

192 

3301 

3527 

3753 

3979 

4205 

4431 

4656 

4882 

5107 

5332 

226 

193 

5557 

5782 

6007 

6232 

6456 

6681 

6905 

7130 

7354 

7578 

225 

194 

7802 

802b 

8249 

8473 

8696 

8920 

9143 

9366 

9589 

9812 

223 

195 

290035 

0257 

0480 

0702 

0925 

1147 

1369 

1591 

1813 

2034 

222 

196 

2256 

2478 

2699 

2920 

3141 

3363 

3584 

3804 

4025 

4246 

221 

197 

4466 

4687 

4907 

5127 

5347 

5567 

5787 

6007 

6226 

6446 

220 

198 

6665 

6884 

7104 

7323 

7542 

7761 

7979 

8198 

8416 

8635 

219 

199 
200 

8853 

9071 
1247 

9289 
1464 

9507 
1681 

9725 
1898 

9943 

.161 

.378 

.595 

2764 

.813 
2980 

218 
217 

301030 

2114 

2331 

2547 

201 

3196 

3412 

3628 

3844 

4059 

4275 

4491 

4706 

4921 

5136 

216 

202 

5351 

5566 

5781 

5996 

6211 

6425 

6639 

6854 

7068 

7282 

215 

203 

7496 

7710  7924 

8137 

8351 

8564 

8778 

8991 

9204 

9417 

213 

204 

9630 

9843 

..56  ,268 

.481 

.693 

.906 

1118 

1330 

1542 

212 

205 

311754 

1966 

2177  2389 

2600 

2812 

3023 

3234 

3445 

3656 

211 

206 

3867 

4078 

4289  4499 

4710 

4920 

5130 

5340 

5551 

5760 

210 

207 

6970 

6180 

6390 

6599 

6809 

7018 

7227 

7436 

7646 

7854 

209 

208 

8063 

8272 

8481 

8689 

8898 

9106 

9314 

9522 

9730 

9938 

208 

209 
210 

320146 

0354 
2426 

0562 
2633 

0769 
2839 

0977 
3046 

1184 
3252 

1391 
3458 

1598 
3665 

1805 
3871 

2012 
4077 

207 
206 

322219 

211 

4282 

4488 

4694 

4899 

5105 

5310 

5516 

5721 

5926 

6131 

205 

212 

6336 

6541 

6745 

6950 

7155 

7359 

7563 

7767 

7972 

8176 

204 

213 

8380 

8583 

8787 

8991 

9191 

9398 

9601 

9805 

...8 

.211 

203 

214 

330414 

0617 

0819 

1022 

1225 

1427 

1630 

1832 

2034 

2236 

202 

215 

2438 

2040 

2842 

3044 

3248 

3447 

3649 

3850 

4051 

4253 1 202  1 

216 

44541  4655 

4856 

5057 

5257 

5458 

5658 

5859 

6059 

6260 

201 

217 

6460  66601 6860 

7060 

7260 

7459 

7659 

7858 

8058 

8257 

200 

218 

84561  8656  8855 

9054 

9253 

9451 

9650 

9849 

..47 

.246 

199 

219 

340444'  0642"  0841'  1039'  1237'  1435'  L632 

1830  2028' 2225'  198  | 

N.  1   0   1  1  1  2  1  3  1  4  1  5  j  6  1  7  1  8  i  9  i  D.  1 

4 

A  TABLE  OP  LOGARITHMS  FROM  1  TO  10,000. 

N.  1   0   |lf2|3|4|5|6|7|8|9|D.  1 

220 
221 
222 
223 
224 
225 
226 
227 
228 
229 
230 
231 
232 
233 
234 
235 
236 
237 
238 
239 
240 
241 
242 
243 
244 
245 
246 
247 
248 
249 
260 
251 
252 
253 
254 
255 
256 
257 
258 
259 
260 
261 
262 
263 
264 
265 
266 
267 
268 
269 
270 
271 
272 
273 
274 
275 
276 
277 
278 
279 

342423 
4392 
6353 
8305 

350248 
2183 
4108 
6026 
7935 
9835 

,2620 
4589 
6549 
8500 
0442 
2375 
4301 
6217 
8125 
..25 
1917 
3800 
5675 
7542 
9401 
1253 
3096 
4932 
6759 
8580 
0392 
2197 
3995 
5785 
7568 
9343 
1112 
2873 
4627 
6374 

2817 
4786 
6744 
8694 
0636 
2568 
4493 
6408 
8316 
.215 
2105 
3988 
5862 
7729 
9687 
1437 
3280 
5115 
6942 
8761 

3014 
4981 
6939 

8889 
0829 
2761 
4685 
6599 
8606 
.404 
2294 
4176 
6049 
7916 
9772 
1622 
3464 
5298 
7124 
8943 

3212 
5178 
7135 
9083 
1023 
2954 
4876 
6790 
8696 
.593 
2482 
4363 
6236 
8101 
9958 
1806 
3G47 
5481 
7306 
9124 

3409 
5374 
7330 

9278 
1216 
3147 
6068 
6981 
8886 
.783 
2671 
4551 
6423 
8287 
.143 
1991 
3831 
5664 
7438 
9306 
1115 
2917 
4712 
6499 
8279 
..51 
1817 
3575 
6326 
7071 
8808 
.538 
2261 
3978 
5688 
7391 
9087 
.777 
2461 
4137 
5808 
7472 
9129 
.781 
2426 
4065 
5697 
7324 
8944 
.559 
2167 
3770 
5367 
6957 
8542 
.122 
1695 
3263 
4825 
6382 

3606 
5570 
7525 
9472 
1410 
3339 
5260 
7172 
9076 
.972 
2859 
4739 
6610 
8473 
.328 
2175 
4015 
5846 
7670 
9487 
1296 
3097 
4891 
6677 
8456 
.228 
1993 
3751 
6501 
7245 
8981 
.711 
2433 
4149 
5868 
7661 
9257 
.946 
2629 
4305 
5974 
7833 
9295 
.945 
2590 
4228 
5860 
7486 
9106 
.720 
2328 
3930 
5526 
7116 
8701 
.279 
1852 
3419 
4981 
6537 

3802 
5766 
7720 
9666 
1603 
3532 
6452 
7363 
9266 
1161 
3048 
4926 
6796 
8659 
.513 
2360 
4198 
6029 
7852 
9668 
1476 
3277 
5070 
6856 
8634 
.405 
5169 
3926 
6676 
7419 
9154 
.883 
2605 
4320 
6029 
7731 
9426 
1114 
2796 
4472 
6141 
7804 
9460 
1110 
2754 
4392 
6023 
7648 
9268 
.881 

3999 
6962 
7915 
9860 
1796 
3724 
5643 
7554 
9456 
1350 
3236 
5113 
6983 
8845 
.698 
2544 
4382 
6212 
8034 
9849 
1656 
3456 
5249 
7034 
8811 
.682 
2345 
4101 
5850 
7592 

4196 
6157 
8110 
..54 
1989 
3916 
6834 
7744 
9646 
1539 
3424 
5301 
7169 
9030 
.883 
2728 
4565 
6394 
8216 
..30 

197 
196 
196 
194 
193 
193 
192 
191 
190 
189 
188 
188 
187 
186 
186 
184 
184 
183 
182 
181 
181 
180 
179 
178 
178 
177 
176 
176 
175 
174 
173 
173 
172 
171 
171 
170 
169 
169 
168 
167 
167 
166 
165 
166 
164 
164 
163 
162 
162 
161 
161 
160 
159 
159 
158 
158 
157 
157 
156 
155 

361728 
3612 
5488 
7356 
9216 

371068 
2912 
4748 
6577 
8398 

380211 
2017 
3815 
5606 
7390 
9166 

390935 
2697 
4452 
6199 

0573 
2377 
4174 
5964 
7746 
9520 
1288 
3048 
4802 
6548 
8287 
..20 
1745 
3464 
5176 
6881 
8579 
.271 
1956 
3635 
5307 
6973 
8633 
.286 
1933 
3574 
5208 
6S36 
8459 
..75 
1686 
3290 
4888 
6481 
8067 
9648 
1224 
2793 
4357 
5915 

0754 
2567 
4353 
6142 
7923 
9698 
1464 
3224 
4977 
6722 
8461 
.192 
1917 
3635 
6346 
7051 
8749 
.440 
2124 
3803 
5474 
7139 
8798 
.451 
2097 
3737 
5371 
6999 
8621 
.236 
1846 
3450 
5048 
6G40 
8226 
9806 
1381 
2950 
4513 
6071 

0934 
2737 
4533 
6321 
8101 
9875 
1641 
3400 
5152 
6896 

1837 
3636 
5428 
7212 
8989 
.769 
2521 
4277 
6025 
7766 

397940 
9674 

401401 
3121 
4834 
6540 
8240 
9933 

411620 
3300 

8114 
9847 
1573 
3292 
6005 
6710 
8410 
.102 
1788 
3467 
5140 
6807 
8467 
.121 
1768 
3410 
5046 
6674 
8297 
9914 
1525 
3130 
4729 
6322 
7909 
9491 
1066 
?637 
4201 
5760 

8634 
.365 
2089 
3807 
65.7 
7221 
8918 
.609 
2293 
3970 
5641 
7306 
8964 
.616 
2261 
3901 
5534 
7161 
8783 
.398 
2007 
3610 
5207 
6798 
8384 
9964 
1538 
3106 
4669 
6226 

9328 
1056 
2777 
4492 
6199 
7901 
9595 
1283 
2964 
4639 

9501 
1228 
2949 
4663 
6370 
8070 
9764 
1451 
3132 
4806 

414973 
6641 
8301 
9956 

421604 
3246 
4882 
6511 
8135 
9752 

6308 
7970 
9625 
1275 
2918 
4565 
6186 
7811 
9429 
1042 
2049 
4249 
5844 
7433 
9017 
.594 
2166 
3732 
5293 
6848 

6474 
8135 
9791 
1439 
3082 
4718 
6349 
7973 
9591 
1203 
2809 
4409 
6004 
7592 
9175 
.752 
2323 
3889 
5449 
7003 

431364 
2969 
4569 
6163 
7751 
9333 

440909 
2480 
4045 
5604 

2488 
4090 
5685 
7275 
8859 
.437 
•2009 
3676 
5137 
6692 

1  N.  1   0   1  1  1  2  1  3  1  4  1  5  1  6  1  7  1  8  1  9  1  D.  1 

A  TABLE  OF  LOGARITHMS  FROM  1 

TO  10,000. 

6 

N. 

1   0   |ll2|3|4|5l6|7!8|9|D.  1 

280 

447158 

7313 

7468 

7623 

7778 

7933 

8088 

8242|  8397 

8552 

155 

281 

8706 

8861 

9015 

9170 

9324 

9478 

9633 

9787 

9941 

..95 

164 

282 

450249 

0403 

0557 

0711 

0865 

1018 

1172 

1326 

1479 

1633 

164 

283 

1786 

1940 

2093 

2247 

2400 

2553 

2706 

2869 

3012 

3165 

153 

284 

3318 

3471 

3624 

3777 

3930 

4082 

4235 

4387 

4540 

4692 

153 

285 

4845 

4997 

5150 

5302 

54.54 

5606 

5758 

6910 

6062 

6214 

152 

286 

6366 

6518 

6670 

6821 

6973 

7125 

7276 

7428 

7579 

7731 

162 

287 

7882 

8033 

8184 

8336 

8487 

8638 

8789 

8940 

9091 

9242 

151 

288 

9392 

9543 

9694 

9845 

9995 

.146 

.296 

.447 

.597 

.748 

151 

289 
290 

460898 
462398 

1048 

2548 

1198 
2697 

1348 

2847 

1499 
2997 

1649 

1799 
3296 

1948 
3446- 

2098 
3594 

2248 
3744 

150 
150 

3146 

291 

3893 

4042 

4191 

4340 

4490 

4639 

4788 

4936 

6086 

6234 

149 

292 

5383 

5532 

5680 

5829 

5977 

6126 

6274 

6423 

6571 

6719 

149 

293 

6868 

7016 

7164 

7312 

7460 

7608 

7756 

7904 

8052 

8200 

148 

294 

8347 

8495 

8643 

8790 

8938 

9085 

9233 

9380 

9527 

9675 

148 

295 

9822 

9969 

.116 

.263 

.410 

.557 

.704 

.851 

.998 

1145 

147 

296 

471292 

1438 

1585 

1732 

1878 

2025 

2171 

2318 

2464 

2610 

146 

297 

2756 

2903 

3049 

3195 

3341 

3487 

3633 

3779 

3925 

4071 

146 

298 

4216 

4362 

4508 

4653 

4799 

4944 

6090 

5235 

5381 

6626 

146 

299 
300 

5671 

5816 

5962 
7411 

6107 
7555 

6253 
7700 

^397 

7844 

6542 

6687 

6832 

8278 

6976 

8422 

145 
145 

477121 

7266 

7989 

8133 

301 

8566 

8711 

8855 

8999 

9143 

9287 

9431 

9575 

9719 

9863 

144 

302 

480007 

0151 

0294 

0^i38 

0582 

0725 

0869 

1012 

1156 

1299 

144 

303 

1443 

1586 

1729 

1872 

2016 

2159 

2302 

2445 

2588 

2731 

143 

304 

2874 

3016 

3159 

3302 

3445 

3587 

3730 

3872 

4015 

4157 

143 

305 

4300 

4442 

4585 

4727 

4869 

5011 

5153 

55295 

5437 

5579 

142 

306 

5721 

5863 

6005 

6147 

6289 

6430 

6572 

6714 

6855 

6997 

142 

307 

7138 

7280 

7421 

7663 

7704 

7845 

7986 

8127 

8269 

8410 

141 

308 

8551 

8692 

8833 

8974 

9114 

9255 

9396 

9537 

9677 

9818 

141 

309 
310 

9958 
491362 

..99 
1502 

.239 

.380 

.520 
1922 

.661 
2062 

.801 
2201 

.941 
2341 

1081 

2481 

1222 
2621 

140 
140 

1642 

1782 

311 

2760 

2900 

3040 

3179 

3319 

3458 

3597 

3737 

3876 

4015 

139 

312 

4155 

4294 

4433 

4572 

4711 

4850 

4989 

6128 

5267 

5406 

139- 

313 

5544 

5083 

5822 

5960 

6099 

6238 

6376 

6515 

6653 

6791 

139 

314 

6930 

7068 

7206 

7344 

7483 

7621 

7759 

7897 

8035 

8173 

138 

315 

8311 

8448 

8586 

8724 

8862 

8999 

9137 

9275 

9412 

9650 

138 

316 

9687 

9824 

9062 

..99 

.236 

.374 

.511 

.648 

.785 

.922 

137 

317 

501059 

1190 

1333 

1470 

1607 

1744 

1880 

2017 

2154 

2291 

137 

318 

2427 

2564 

2700 

2837 

2973 

3109 

3246 

3382 

3518 

3655 

136 

319 
320 

3791 
505150 

3927 
5286 

4063 
5421 

4199 
5557 

4335 
5693 

4471 

5828 

4607 

4743 
6099 

4878 
6234 

5014 
6370 

136 
136 

5964 

321 

6505 

6640 

6776 

6911 

7046 

7181 

7316 

7451 

7586 

7721 

135 

322 

7856 

7091 

8120 

8260 

8395 

8530 

8664 

8799 

8934 

9068 

135 

323 

9203 

9337 

9471 

9606 

9740 

9874 

...9 

.143 

.277 

.411 

134 

324 

510545 

0679 

0813 

0947 

1081 

1215 

1349 

1482 

1616 

1750 

134 

325 

1883 

2017 

2151 

2284 

2418 

2551 

2684 

2818 

2951 

3084 

133 

326 

3218 

3351 

3484 

3617 

3750 

3883 

4016 

4149 

4282 

4414 

133 

327 

4548 

4681 

4813 

4946 

5079 

5211 

5344 

5476 

5609 

5741 

133 

328 

5874 

6000 

6139 

6271 

6403 

6535 

6668 

6800 

6932 

7064 

132 

329 

7196 

7328 

7460 

7592 

7724 

7855 

7987 

8119 

8251 

8382 

132 

330 

518514 

8646 

8777 

8909 

9040 

9171 

9.303 

9434 

9566 

9697 

131 

331 

9828 

9959 

..90 

.221 

..353 

.484 

.615 

.745 

.876 

1007 

131 

332 

521138 

1269 

1400 

1530 

1661 

1792 

1922 

2053 

2183 

2314 

131 

333 

2444 

2575 

2705 

2835 

2966 

3096 

3226 

3356 

3486 

3616 

130 

334 

3746 

8876 

4006 

4136 

4266 

4396 

4526 

4656 

4785 

4915 

130 

335 

5045 

5174 

5304 

5434 

5563 

6693 

5822 

6951 

6081 

6210 

129 

336 

6339 

6469 

6598 

6727 

6856 

6985 

7114 

7243 

7372 

7601 

129 

337 

7630 

7759 

7888 

8016 

8145 

8274 

8402 

8531 

8660 

8788 

129 

338 

8917 

9045 

9174 

9302 

9430 

9559 

9687 

9815 

9943 

..72 

128 

339 

530200 

0328  0456 

0584 

0712  0840 

0968  1096 

1223 

1361 

128 

"nT 

0   |l|2|3l4|5|6l7|8|9|D.  1 

A  TABLE  OF  LOGARITHMS  rXOM  1  TO  10,000 


N. 

1   0   |l|2|3|4|5|6|7|8|9|D.  1 

340 

531479! 1607 

1734 

1862 

1990 

2117 

2245 

2372 

2500 

2627 

128 

341 

2754 

2882 

3009 

3136 

3264 

3391 

3518 

3645 

3772 

.3899 

127 

342 

4026 

4153 

4280 

4407 

4534 

4661 

4787 

4914 

,5041 

5167 

127 

343 

5294 

5421 

5547 

5674 

5800 

5927 

6053 

6180 

6306 

6432 

126 

344 

6558 

6685 

6811 

6937 

7063 

7189 

7315 

7441 

7567 

7693 

126 

345 

7819 

7945 

8071 

8197 

8322 

8448 

8574 

8699 

8825 

8951 

126 

346 

9076 

9202 

9327 

9452 

9578 

9703 

9829 

9954 

..79 

.204 

125 

347 

540329 

0455 

0580 

0705 

0830 

0955 i  1080 

1205 

1330 

14.54 

125 

348 

1579 

1704 

1829 

1953 

2078 

2203 

2327 

2452 

2576 

2701 

125 

349 
350 

2825 

2950 
4192 

3074 

3199 
4440 

3323 

3447 

4688 

3571 
4812 

3696 
4936 

.38ro 

.5060 

3944 

124 
124 

544068 

4316 

4564 

5183 

351 

5307 

6431 

5555 

5678 

5802 

5925 

6049 

6172 

6296 

6419 

124 

352 

6543 

6666 

6789 

6913 

7036 

7159 

7282 

7405 

7529 

7652 

123 

853 

7775 

7898 

8021 

8144 

8267 

8389 

8512 

8635 

87,58 

8881 

123 

354 

-  9003 

9126 

9249 

9371 

9494 

9616 

9739 

9861 

9984 

.106 

123 

355 

550228 

0351 

0473 

0595 

0717 

0840 

0962 

1084 

1206 

1.328 

122 

356 

1450 

1572 

1694 

1816 

1938 

2060 

2181 

2303 

2425 

2547 

122 

357 

2668 

2790 

2911 

3033 

3155 

3276 

,3398 

3519 

3640 

3762 

121 

358 

3883 

4004 

4126 

4247 

4368 

4489 

4610 

4731 

4852 

4973 

121 

359 
360 

5094 

5215 

5336 

5457 
6664 

5578 
6785 

5699 
6905 

5820 

5940 

6061 
7267 

6182 
7,387 

121 
120 

556303 

6423 

6544 

7026 

7146 

361 

7507 

7627 

7748 

7868 

7988 

8108 

8228 

8349 

8469 

8589 

120 

362 

8709 

8829 

8948 

9068 

9188 

9308 

9428 

9548 

9667 

9787 

120 

363 

9907 

..26 

.146 

.265 

.385 

.504 

.624 

.743 

.863 

.982 

119 

364 

561101 

1221 

1340 

1459 

1578 

1698 

1817 

1936 

2055 

2174 

119 

365 

2293 

2412 

2531 

2650 

2769 

2887 

3006 

3125 

3244 

3362 

119 

366 

3481 

3600 

3718 

3837 

3955 

4074 

4192 

4311 

4429 

4548 

119 

367 

4666 

4784 

4903 

5021 

5139 

5257 

5376 

,5494 

.5612 

5730 

118 

368 

5848 

5966 

6084 

6202 

6320 

6437 

6555 

6673 

6791 

6909 

118 

369 
370 

7026 
568202 

7144 
8319 

7262 
8436 

7379 

8554 

7497 
8671 

7614 

8788" 

7732 
8905 

7849 
902.'^ 

7967 

8084 
9257 

118 
117 

9140 

371 

9374 

9491 

9608 

9725 

9842 

9959 

..76 

.19o 

.309 

.426 

117 

372 

570543 

0660 

0776 

0893 

1010 

1126 

1243 

1 359 

1476 

1.592 

117 

373 

1709 

1825 

1942 

2058 

2174 

2291 

2407 

2523 

26.39 

2755 

116 

374 

2872 

2988 

3104 

3220 

3336 

3452 

3568 

3684 

3800 

3915 

116 

375 

4031 

4147 

4263 

4379 

4494 

4610 

4726 

4841 

4957 

.5072 

116 

376 

5188 

5303 

5419 

5534 

5650 

5765 

5880 

5996 

6111 

6226 

115 

377 

6341 

6457 

6572 

6687 

6802 

6917 

7032 

7147 

7262 

7377 

115 

378 

7492 

7607 

7722 

7836 

7951 

8066 

8181 

8295 

8410 

8525 

115 

379 
380 

8639 

8754 
9898 

8868 

8983 
.126 

9097 
.241 

9212 
.355 

9326 
.469 

9441 

9555 

9669 
.811 

114 
114 

579784 

..12 

.583 

.697 

381 

580925 

1039 

1153 

1267 

1381 

1495 

1608 

1722 

1836 

19,50 

114 

382 

2063 

2177 

2291 

2404 

2518 

2631 

2745 

28.58 

2972 

3085 

114 

383 

3199 

3312 

3426 

3539 

3652 

3705 

.3879 

3992 

4105 

4218 

113 

384 

4331 

4444 

4557 

4670 

4783 

4896 

5009 

5122 

5235 

5348 

113 

385 

5461 

6574 

5686 

5799 

5912 

6024 

6137 

62,50 

6362 

6475 

H3 

386 

6587 

6700 

6812 

6925 

7037 

7149 

7262 

7374 

7486 

7599 

112 

387 

7711 

7823 

7935 

8047 

8160 

8272 

8384 

8496 

8608 

8720 

112 

388 

8832 

8944 

9056 

9167 

9279 

9391 

9503 

9615 

9726 

9838 

112 

389 
590 

9950 

..61 
1176 

.173 

.284 
1399 

.396 
1510 

.507 
1021 

.619 

.730 

.842 
1 955 

.953 
2066 

112 
111 

591065 

1287 

1732 

1843 

391 

2177 

2288 

2399 

2510 

2621 

2732 

2843 

2954 

3064 

3175 

111 

392 

3286 

3397 

3508 

3618 

3729 

3840 

3950 

4061 

4171 

4282^ 

111 

393 

4393 

4503 

4614 

4724 

4834 

4945 

5055 

5165 

.5276 

5386 

110 

394 

5496 

5606 

5717 

5827 

5937 

6047 

61.57 

6267 

6377 

6487 

110 

395 

6597 

6707 

6817 

6927 

7037 

7146 

7256 

7366 

7476 

7586 

110 

396 

7695 

7805 

7914 

8024 

8134 

8243 

8.353 

8462 

8572 

8681 

110 

397 

8791 

8900 

9009 

9119 

9228 

9337 

9446 

95.56 

9665 

9774 

109 

398 

9883 

9992 

.101 

.210 

.319 

.428 

..537 

.646 

.7.55 

.864 

109 

399 

600973 

1082 

1191 

1299 

1408 

1517 

1625 

1734 

1843 

1951 

109 

N. 

0   |l|2|3|4|5i6|7|8|9|D.  1 

I 


A  TABLE  OF  LOGARITHMS  FROM  1  TO  10,000. 

7 

N.  1   0   1  1  1  2  1  3  !  4  1  5  1  6  1  7  !  8  1  9  1  D  1 

400 
401 
402 
403 
404 
405 
406 
407 
408 
409 

4ro 

411 
412 
413 
414 
415 
416 
417 
418 
419 
420 
421 
422 
423 
424 
425 
426 
427 
428 
4','.9 
4S0 
431 
432 
433 
434 
435 
436 
437 
438 
439 
440 
441 
442 
443 
444 
445 
446 
447 
44S 
449 
450 
451 
452 
453 
454 
455 
456 
457 
458 
459 

602060 
3144 
4226 
5305 
6381 
7455 
8526 
9594 

610660 
1723 

2169 
3253 
4334 

5413 
6489 
7562 
8633 
9701 
0767 
1829 
2890 
3947 
5003 
6055 
7105 
8153 
9198 
0240 
1280 
2318 
3353 
4385 
5415 
6443 
7468 
8491 
9512 
0530 
1545 
2559 
3569 
4578 
5584 
6588 
7590 
8589 
9586 
0581 
1573 
2563 
3551 
4537 
5521 
6502 
7481 
8458 
9432 
0405 
1375 
2343 
3309 
4273 
5235 
6194 
7152 
8107 
9060 
..11 
0960 
]  907 

2277 
3361 
4442 
5521 
6596 
7669 
8740 
9808 
0873 
1936 
2996 
4053 
5108 
6160 
7210 
8257 
9302 
0344 
1384 
2421 
3456 
4488 
5518 
6546 
7571 
8593 
9613 
0631 
1647 
2660 
3670 
4679 
5685 
6688 
7690 
8689 
9686 
0680 
1672 
2662 
3650 
4636 
5619 
6600 
7579 
8555 
9530 
0502 
1472 
2440 

3405 
4369 
5331 
6290 
7247 
8202 
9155 
.106 
1055 
2002 

2386 
3469 
4550 
5628 
6704 
7777 
8847 
9914 
0979 
2042 
3102 
4159 
5213 
6265 
7315 
8362 
9106 
0448 
1488 
2525 
3559 
4591 
5621 
6648 
7673 
8695 
9715 
0733 
1748 
2761 
3771 
4779 
5785 
6789 
7790 
8789 
9785 
0779 
1771 
2761 
3749 
4734 
5717 
6698 
7676 
8653 
9627 
0599 
1569 
2536 

3502 
4465 
5427 
6386 
7343 
8298 
9250 
.201 
1 150 
2096 

2494 
3577 

4658 
5736 
6811 
7884 
8954 
..21 
1086 
2148 
3207 
4264 
5319 
6370 
7420 
8466 
9511 
0552 
1592 
2628 
3663 
4695 
5724 
6751 
7775 
8797 
9817 
0835 
1849 
2862 
3872 
4880 
5886 
6889 
7890 
8888 
9885 
0879 
1871 
2860 
3847 
4832 
5815 
6796 
7774 
8750 
9724 
0696 
1666 
2633 
3598 
4562 
5523 
6482 
7438 
8393 
9346 
.296 
1245 
2191 

2603 
3686 
4766 
5844 
6919 
7991 
9061 
.128 
1192 
2254 
3313 
4370 
5424 
6476 
7525 
8571 
9615 
0656 
1695 
2732 
3766 
4798 
5827 
6853 
7878 
8900 
9919 
0936 
1951 
2963 
3973 
4981 
5986 
&989 
7990 
8988 
9984 
0978 
1970 
2959 
3946 
4931 
5913 
6894 
7872 
8848 
9821 
0793 
1762 
2730 
3695 
4658 
5619 
6577 
7534 
8488 
9441 
.391 
1339 
2280 

2711 
3794 
4874 
5951 
7026 
8098 
9167 
.234 
1298 
2360 

3419 
4475 
5529 
6581 
7629 
8676 
9719 
0760 
1799 
2835 

3869 
4901 
5929 
6956 
7980 
9002 
..21 
1038 
2052 
3064 
4074 
5081 
6087 
7089 
8090 
9088 
..84 
1077 
2069 
30.58 
4044 
5029 
6011 
6992 
7969 
8945 
9919 
0890 
1859 
2826 

3791 
4754 
5715 
6673 
7629 
8584 
9536 
.486 
1434 
2380 

2819 
3902 
4982 
6059 
7133 
8205 
9274 
.341 
1405 
2466 
3525 
4581 
5634 
6686 
7734 
8780 
9824 
0864 
1903 
2939 
3973 
5004 
6032 
7058 
8082 
9104 
.123 
1139 
2153 
3165 
4175 
5182 
6187 
7189 
8190 
9188 
.183 
1177 
2168 
3156 
4143 
5127 
6110 
7089 
8067 
9043 
..16 
0987 
1956 
2923 
3888 
4850 
.5810 
6769 
7725 
8679 
9631 
.581 
1529 
2475 

29281 
4010 
5089 
6166 
7241 
8312 
9381 
.447 
1511 
2572 
3630 
4686 
5740 
6790 
7839 
8884 
9928 
0968 
2007 
3042 
4076 
5107 
6135 
7161 
8185 
9206 
.224 
1241 
2255 
3266 
4276 
5283 
6287 
72{i0 
8290 
9287 
.283 
1276 
2267 
3255 
4242 
5226 
6208 
7187 
8165 
9140 
.113 
1084 
2053 
3019 
3984 
4946 
5906 
6864 
7820 
8774 
9726 
.676 
1623 
2569 

3036 

4118 
5197 
6274 
7348 
8419 
9488 
.554 
1617 
2678 
3736 
4792 
5845 
6895 
7943 
8989 
..32 
1072 
2110 
3146 
4179 
5210 
6238 
7263 
8287 
9308 
.326 
1342 
2356 
3367 
4376 
5383 
6388 
7390 
8389 
9387 
.382 
1375 
2366 
3354 
4340 
5324 
6306 
7285 
8262 
9237 
.210 
1181 
2150 
3116 
4080 
5042 
6002 
6960 
7916 
8870 
9821 
.771 
1718 
2663 

108 

108 

108 

108 

107 

107 

107 

107 

106 

106 

106 

106 

105 

105 

105 

105 

104 

104 

104 

104 

\03 

103 

103 

103 

102  ' 

103 

102 

102 

101 

101 

100 

100 

100 

WO 

99 

99 

99 

99 

99 

99 

98 

98 

98 

98 

98 

97 

97 

97 

97 

97 

96 

96 

96 

96 

96 

95 

95 

95 

95 

95 

612784 
3842 
4897 
5950 
7000 
8048 
9093 

620136 
1176 
2214 

623249 
4282 
5312 
6340 
7366 
8389 
9410 

630428 
1444 
2457 

633468 
4477 
5484 
6488 
7490 
8489 
9486 

640481 
1474 
2465 

643453 
4439 
5422 
6404 
7383 
8360 
9335 

650308 
1278 
2246 

653213 
4177 
5138 
6098 
7056 
8011 
8965 
9916 

660865 
1813 

N.  !   (»   1  1  1  2  1  3  t  4  1  5  1  6  1  7  1  8  1  9  !  D.  1 

« 

A  TABLE  OF  LOGARITHMS  FROM  1  TO  10,000, 

N. 

0   ;ii2|3|4|5|6|7|8|9|D.  1 

460 

662758 

28521 2947 

3041  3135 1  3230,  3324,  3418] 

3512  3607,  941 

461 

3701 

3795 

3889 

3983  40781 

4172  4266  4360 

4464 

4548 

94 

462 

4642 

4736 

4830 

4924 

5018 

5112  6206] 

5299 

5393 

,5487 

94 

463 

6581 

5675 

5769 

6862 

5956 

6050 

6143 

6237 

6331 

6424 

94 

464 

6518 

6612 

6705 

6799 

6892 

6986 

7079 

7173 

7266 

7360 

94 

i65 

7463 

7546 

7640 

7733 

7826 

7920 

8013 

8106 

8199 

8293 

93 

466 

8386 

8479 

8572 

8665 

8759 

8852 

8945 

9038 

9131 

9224 

93 

467 

9317 

9410 

9503 

9596 

9689 

9782 

9875 

9967 

..60 

.1,53 

93 

468 

670246 

0339 

0431 

0524 

0617 

0710 

0802 

0895 

0988 

1080 

93 

469 

470 

1173 

1265 
2190 

1368 

1451 

1543 

1636 
2660 

1728 
2652 

1821 
2744 

1913 

2005 
2929 

93 
92 

672098 

2283 

2375 

2467 

2836 

471 

3021 

3113 

3205 

3297 

3390 

3482 

3574 

3666 

3758 

3850 

92 

472 

3942 

4034 

4126 

4218 

4310 

4402 

4494 

4586 

4677 

4769 

92 

473 

4861 

4953 

5045  5137 

5228 

6320 

.5412 

5603 

6695 

5687 

92 

474 

5778 

5870 

5962 

6053 

6145 

6236 

6328 

6419 

6511 

6602 

92 

475 

6694 

6785 

6876 

6968 

7059 

7151 

7242 

7333 

7424 

7516 

91 

476 

7607 

7698 

7789 

7881 

7972 

8063 

8164 

8246 

8336 

8427 

91 

477 

8518 

8609 

8700 

8791 

8882 

8973 

9064 

9155 

9246 

9337 

91 

478 

9428 

9519 

9610 

9700 

9791 

9882 

9973 

..63 

.154 

.245 

91 

479 

480 

680336 
681241 

0426 
1332 

0517 

0607 

0698 
1603 

0789 
1693 

0879 
1784 

0970 

1874 

1060 

1151 
2055 

91 
90 

1422 

1513 

1964 

481 

2145 

2235 

2326 

2416 

2506 

2596 

2686 

2777 

2867 

2967 

90 

482 

3047 

3137 

3227 

3317 

3407 

3497 

3587 

3677 

3767 

3857 

90 

483 

3947 

4037 

4127 

4217 

4307 

4396 

4486 

4676 

4666 

4756 

90 

484 

4845 

4935 

6025 

6114 

5204 

5294 

5383 

5473 

5563 

5652 

90 

485 

6742 

5831 

5921 

6010 

6100 

6189 

6279 

6368 

6458 

6547 

89 

486 

6036 

6726 

6815 

6904 

6994 

7083 

7172 

7261 

7351 

7440 

89 

487 

7629 

7618 

7707 

7796 

7886 

7975 

8064 

8153 

8242 

8331 

89 

488 

8420 

8509 

8598 

8687 

8776 

8865 

8953 

9042 

9131 

9220 

89 

489 
490 

9309 
690196 

9398 
0285 

9486 

9575 

9664 

9753 

9841 

9930 

.,19 

.107 

89 
89 

0373 

0462 

0550 

0639 

0728 

0816 

0905 

0993 

491 

1081 

1170 

1258 

.  1347 

1435 

1524 

1612 

1700 

1789 

1877 

88 

492 

1965 

2053 

2142 

2230 

2318 

2406 

2494 

2583 

2671 

2759 

88 

493 

2847 

2935 

3023 

3111 

3199 

3287 

3375 

3463 

3551 

3639 

88 

494 

3727 

3815 

3903 

3991 

4078 

4166 

4254 

4342 

4430 

4517 

88 

495 

4605 

4693 

4781 

4868 

4956 

5044 

6131 

5219 

5307 

5394 

88 

496 

6482 

5569 

5657 

5744 

5832 

.5919 

6007 

6094 

6182 

6269 

87 

497 

6356 

6444 

6631 

6618 

6706 

6793 

6880 

6968 

7055 

71421  87  1 

498 

7229 

7317 

7404 

7491 

7578 

7665 

7752 

7839 

7926 

8014 

87 

499 
500 

8101 
698970 

8188 

8275 
9144 

8362 

8449 
9317 

8535 
9404 

8622 
9491 

8709 
9578 

8796 
9664 

8883 
9751 

87 
87 

9057 

9231 

501 

9838 

9924 

..11 

..98 

.184 

.271 

.3.58 

,444 

.531 

.617 

87 

502 

700704 

0790 

0877 

0963 

1050 

1136 

1222 

1309 

1395 

1482 

86 

503 

1668 

1654 

1741 

1827 

1913 

1999 

2086 

2172 

2258 

2344 

86 

504 

2431 

2517 

2603 

2689 

2775 

2861 

2947 

3033 

3119 

3205 

86 

505 

3291 

3377 

3463 

3549 

3635 

3721 

3807 

,3895 

3979 

4065 

S6 

506 

4151 

4236 

4322 

4408 

4494 

4679 

4665 

4761 

4837 

4922 

86 

507 

6008 

5094 

5179 

6265 

6350 

.5436 

5522 

5607 

5693 

5778 

86 

508 

6864 

5949 

6036 

6120 

6206 

6291 

6376 

6462 

6547 

66,32 

85 

509 
510 

6718 
707670 

6803 

6888 
7740 

6974 

7059 
7911 

7144 
799ri 

2229 
8081 

7315 
8166 

7400 

8251 

7485 
8,336 

85 
85 

7665 

7826 

511 

8421 

8506 

8591 

8676 

8761 

8846 

8931 

9015 

9100 

9185 

85 

512 

9270 

9355 

9440 

9524 

9609 

9694 

9779 

9863 

9948 

..33 

85 

513 

710117 

0202 

0287 

0371 

0456 

0540 

0625 

0710 

0794 

0879 

85 

514 

0963 

1048 

1132 

1217 

1301 

1385 

1470 

1.554 

1639 

1723 

84 

515 

1807 

1892 

1976 

2060 

2144 

2229 

2313 

2397 

2481 

2566 

84 

516 

2650 

2734 

2818  2902 

2986 

3070 

3154 

3238 

3323 

3407 

84 

517 

3491 

3575 

3650  3742 

3826 

,3910 

3994 

4078 

4162 

4246 

84 

518 

4330 

4414 

4497  4581 

4665 

4749 

4833 

4916 

5000 

5084 

84 

519 

5167' 5251 

5335  5418 

65021  55861  56691  5753'  5836'  5920 

84 

N. 

1   0   1  1  1  2  1  3  1  4  1  5  1  6  1  7  1  8  1  9  1  D.  1 

A  TABLE  OF  LOGARITHMS  FROM  1 

TO  10,000 

9 

N. 

1   0   |l|2|3|4|5|6|7|8|9|D.  1 

620 

7160031  6087i  6170 

6254 

6337 

6421 

6504 

6688 

6671 

6754 

83 

521 

6838 

6921 

7004 

7088 

7171 

7254 

7338 

7421 

7504 

7587 

83 

fi22 

7671 

7754 

7837 

7920 

8003 

8086 

8169 

8253 

8336 

8419 

83 

523 

8502 

8585 

8668 

8751 

8834 

8917 

9000 

9083 

9165 

9248 

83 

524 

9331 

9414 

9497 

9580 

9663 

9745 

9828 

9911 

9994 

..77 

83 

525 

720159 

0242 

0325 

0407 

0490 

0573 

0666 

0738 

0821 

0903 

83 

526 

0986 

1068 

1151 

1233 

1316 

1398 

1481 

1663 

1646 

1728 

82 

527 

1811 

1893 

1975 

2058 

2140 

2222 

2306 

2387 

2469 

2552 

82 

528 

2634 

2716 

2798 

2881 

2963 

3045 

3127 

3209 

3291 

3374 

82 

529 

3456 

3538 

3620 

3702 

3784 

3866 

3948 

4030 

4112 

4194 

82 

530 

724276 

4358 

4440 

4522 

4604 

4685 

4767 

4849 

4931 

5013 

82 

531 

5095 

5176 

6258 

5340 

5422 

6503 

5685 

5667 

5748 

5830 

82 

532 

5912 

5993 

6075 

6156 

6238 

6320 

6401 

6483 

6664 

6646 

82 

533 

6727 

6809 

6890 

6972 

7053 

7134 

7216 

7297 

7379 

7460 

81 

534 

7541 

7623 

7704 

7785 

7866 

7948 

8029 

8110 

8191 

8273 

81 

535 

8354 

8435 

8516 

8597 

8678 

8769 

8841 

8922 

9003 

9084 

81 

53fi 

9165 

9246 

9327 

9408 

9489 

9570 

9651 

9732 

9813 

9893 

81 

537 

9974 

..55 

.136 

.217 

.298 

.378 

.469 

•540 

.621 

.702 

81 

538 

730782 

0863 

0944 

1024 

1105 

1186 

1266 

1347 

1428 

1508 

81 

639 

1589 

1C69 

1750 

1830 

1911 

1991 

2072 

2152 

2233 

2313 

81 

540 

732394 

2474 

2555 

2635 

2716 

2796 

2876 

2956 

3037 

3117 

80 

541 

3197 

3278 

3358 

3438 

3518 

3698 

3679 

3759 

3839 

3919 

80 

542 

3999 

4079 

4160 

4240 

4320 

4400 

4480 

4660 

4640 

4720 

80 

543 

4800 

4880 

4960 

5040 

5120 

6200 

6279 

5359 

5439 

5619 

80 

544 

6599 

5679 

6759 

5838 

5918 

5998 

6078 

6157 

6237 

6317 

80 

545 

6397 

6476 

6556 

6635 

6715 

6795 

6874 

6954 

7034 

7113 

80 

546 

7193 

7272 

7352 

7431 

7511 
8305 

7590 

7670 

7749 

7829 

7908 

79 

547 

7987 

8067 

8146 

8225 

8384 

8463 

8543 

8622 

8701 

79 

548 

8781 

8860 

8939 

9018 

9097 

9177 

9266 

9335 

9414 

9493 

79 

549 
550 

9572 

9651 
0442 

9731 
0521 

9810 
0600 

9889 
0678 

9968 
0757 

..47 
0836 

.126 
0916 

.205 
0994 

.284 
1073 

79 
79 

740363 

551 

1152 

1230 

1309 

1388 

1467 

1546 

1624 

1703 

1782 

1860 

79 

552 

1939 

2018 

2096 

2175 

2254 

2332 

2411 

2489 

2568 

2646 

79 

553 

2725 

2804 

2882 

2961 

3039 

3118 

3196 

3275 

3353 

3431 

78 

554 

3510 

3588 

3667 

3745 

3823 

3902 

3980 

4058 

4136 

4215 

78 

555 

4293 

4371 

4449 

4628 

4606 

4684 

4762 

4840 

4919 

4997 

78 

556 

5075 

6163 

5231 

6309 

5387 

5465 

6543 

5621 

5699 

5777 

78 

557 

5855 

5933 

6011 

6089 

6167 

6245 

6323 

6401 

6479 

6556 

78 

558 

6C34 

6712 

6790 

6868 

6946 

7023 

7101 

7179 

7256 

7334 

78 

569 
560 

7412 

7489 
8266 

7667 
8343 

7645 
8421 

7722 
8498 

7800 
8676 

7878 

7955 

8033 

8808 

8110 

8885 

78 
77 

748188 

8653 

8731 

561 

8963 

9040 

9118 

9195 

9272 

9360 

9427 

9504 

9582 

9659 

77 

562 

9736 

9814 

9891 

9968 

..46 

.123 

,,200 

.277 

.354 

.431 

77 

563 

750508 

0686 

0663 

0740 

0817 

0894 

0971 

1048 

1125 

1202 

77 

564 

1279 

1366 

1433 

1510 

1587 

1664 

1741 

1818 

1895 

1972 

77 

565 

2048 

2125 

2202 

2279 

2356 

2433 

2509 

-2586 

2663 

2740 

77 

566 

2816 

2893 

2970 

3047 

3123 

3200 

3277 

3353 

3430 

3506 

77 

567 

3583 

3660 

3736 

3813 

3889 

3966 

4042 

4119 

4196 

4272 

77 

568 

4348 

4425 

4601 

4578 

4654 

4730 

4807 

4883 

4960 

5036 

76 

569 

5112 

6189 

5266 

6341 

5417 

5494 

6570 

6646 

5722 

5799 

76 

570 

755876 

5951 

6027 

6103 

6180 

6256 

6332 

6408 

6484 

6560 

76 

571 

6636 

6712 

6788 

6864 

6940 

7016 

7092 

7168 

7244 

7320 

76 

572 

7396 

7472 

7548 

7624 

7700 

7775 

7861 

7927 

8003 

8079 

76 

573 

8155 

8230 

8306 

8382 

8458 

8533 

8609 

8685 

8761 

8836 

76 

574 

8912 

8988 

9063 

9139 

9214 

9290 

9366 

9441 

9517 

9592 

76 

575 

9668 

9743 

9819 

9894 

9970 

..45 

.121 

.196 

.272 

..347 

75 

576 

760422 

0498 

0573 

0649 

0724 

0799 

0875 

0950 

1025 

1101 

75 

577 

1176 

1251 

1326 

1402 

1477 

1562 

1627 

1702 

1778 

1863 

75 

578 

1928 

2003 

2078 

21531  2228 

2303 

2378 

2453 

2529; 2604 

75 

579 

2679 

275412829'  2904'  '^\f'i8 

3053  3128' 3203 

327813353'  75  | 

N. 

0   lll2l3|4|5|6|7|8l9lD7| 

10 


A  TABLE  or  LOGARITHMS  FBOM  1  TO  10,000. 


N.  1   0   |lf2|3l4|5|6|7|8|9|D. 

680 

581 

582 

583 

584 

585 

586 

587 

588 

589 

590 

591 

592 

593 

694 

595 

596 

597 

598 

599 

600 

601 

602 

603 

604 

605 

606 

607 

608 

€09 

610 

611 

612 

613 

614 

615 

616 

617 

618 

619 

620 

621 

622 

623 

624 

625 

626 

627 

628 

629 

630 

631 

632 

633 

634 

635 

636 

637 

638 

639 

763426 
4176 
4922 
566S 
6413 
7156 
7898 
8638 
9377 

770115 

I   3502 

4251 

4998 

5743 

6487 

7230 

7972 

8712 

9451 

0189 

0926 

1661 

2395 

3128 

3860 

4590 

5319 

6047 

6774 

7499 

8224 

8947 

9669 

0389 

1109 

1827 

2544 

3260 

3975 

4689 

5401 

6112 

6822 

7531 

8239 

8946 

9651 

0356 

1059 

1761 

2462 

3162 

3860 

4558 

5254 

5949 

6644 

7337 

8029 

8720 

9409 

0098 

0786 

1472 

2158 

2842 

3525 

4208 

4889 

5569 

3576 
4326 
5072 
5818 
6562 
7304 
8046 
8786 
9525 
0263 
0999 
1734 
2468 
3201 
3933 
4663 
5392 
6120 
6846 
7572 
8296 
9019 
9741 
0461 
1181 
1899 
2616 
3332 
4046 
4760 
5472 
6183 
6893 
7602 
8310 
9016 
9722 
0426 
1129 
1831 

2532 
3231 
3930 
4627 
5324 
6019 
6713 
7406 
8098 
8789 
9478 
0167 
0854 
1541 
2226 
2910 
3594 
4276 
4957 
5637 

3653 

4400 

5147 

5892 

6636 

7379 

8120 

8860 

9599 

0336 

1073 

1808 

2542 

3274 

4006 

4736 

5466 

6193 

6919 

7644 

8368 

9091 

9813 

0533 

1253 

1971 

2688 

3403 

4118 

4831 

5543 

6254 

6964 

7673 

8381 

90S7 

9792 

0496 

119& 

1901 

2602 
3301 
4000 
4697 
5393 
6088 
6782 
7475 
8167 
8858 
9547 
0236 
0923 
1609 
2295 
2979 
3662 
4344 
5025 
5705 

3727 
4475 
6221 
5966 
6710 
7453 
8194 
8934 
9673 
0410 

1146 
1881 
2615 
3348 
4079 
4809 
6538 
6265 
6992 
7717 
8441 
9163 
9885 
0605 
1324 
2042 
2759 
3476 
4189 
4902 
5616 
6325 
7035 
7744 
8451 
9167 
9863 
0667 
1269 
1971 

2672 
3371 
4070 
4767 
6463 
6168 
6852 
7546 
8236 
8927 
9616 
0306 
0992 
1678 
2363 
3047 
3730! 
44121 
5093 
6773I 

'  3802 
465C 
5296 
6041 
6785 
7527 
8268 
9008 
9746 
0484 
1220 
1955 
2688 
3421 
4152 
4882 
5610 
6338 
7064 
7789 
8513 
9236 
9967 
0677 
1396 
2114 
2831 
3546 
4261- 
4974 
5686 
6396 
7106 
7815 
8622 
9228 
9933 
0637 
1340 
2041 

2742 
3441 
4139 
4836 
6632 
6227 
6921 
7614 
8305 
8996 
9685 
0373 
1061 
1747 
2432 
3116 
3798 
4480 
5161 
5841 

3877 

4624 

5370 

6115 

6869 

7601 

8342 

9082 

9820 

0657 

1293 

202s 

2762 

3494 

4225 

4955 

5683 

6411 

7137 

7862 

8585 

9308 

..29 

0749 

1468 

2186 

2902 

3618 

4332 

6045 

5757 

6467 

7177 

7885 

8593 

9299 

...4 

0707 

1410 

2111 

2812 

3611 

4209 

4906 

5602 

6297 

6990 

7683 

8374 

9065 

9764 

0442 

1129 

1816 

2500 

3184 

3807 

4548 

5229 

5908 

3952 

4699 

5445 

619( 

6933 

7675 

8416 

9156 

9894 

0631 

1367 

2102 

2835 

3567 

4298 

5028 

5766 

6483 

7209 

7934 

8658 

9380 

.101 

0821 

1540 

2258 

2974 

3689 

4403 

5116 

5828 

6538 

7248 

7956 

8663 

9369 

..74 

0778 

1480 

2181 

2882 

3581 

4279 

4976 

5672 

6366 

7060 

7752 

8443 

9134 

9823 

0511 

1198 

1884 

2568 

3252 

3935 

4616 

52U7 

59^6 

4027 

4774 

5520 

6264 

7007 

7749 

8490 

9230 

9968 

0705 

1440 

2176 

2908 

3640 

4371 

5100 

5829 

6556 

7282 

8006 

8730 

9462 

.173 

0893 

1612 

2329 

3046 

3761 

4475 

5187 

5899 

6609 

7319 

8027 

8734 

9440 

.144 

0848 

1560 

2262 

2952 

3651 

4349 

5045 

5741 

6436 

7129 

7821 

8513 

9203 

9892 

0580 

1266 

1952 

2637 

3321 

4003 

4685 

5365 

6044' 

4101 

4848 

6594 

6338 

7082 

7«23 

8564 

9303 

..42 

0778 

1514 

2248 

2981 

3713 

4444 

5173 

6902 

6629 

7364 

8079 

8802 

9524 

.245 

0966 

1684 

2401 

3117 

3832 

4646 

5269 

5970 

6680 

7390 

8098 

8804 

9510 

.215 

0918 

1620 

2322 

3022 

3721 

4418 

5115 

5811 

6506 

7198 

7890 

8582 

9272 

9961 

0648 

1335 

2021 

2705 

3389 

4071 

4753 

5433 

6112 

75 
75 
75 

74 

74 

74 

74 

74 

74 

74 

74 

73 

73 

73 

73 

73 

73 

73 

73 

72 

72 

72 

72 

72 

72 

72 

72 

71 

71 

71 

71 

71 

71 

71 

71 

71 

70 

70 

70 

70 

70 

70 

70 

70 

70 

69 

69 

69 

69 

69 

69 

69 

69 

69 

69 

68 

68 

68 

68 

68 

770852 
1587 
2322 
3055 
3786 
4517 
5246 
5974 
6701 
7427 

778151 
8874 
9596 

780317 
1037 
1755 
2473 
3189 
3904 
4617 

785330 
6041 
6751 
7460 
8168 
8875 
9581 

790285 
0988 
1691 

792392 
3092 
3790 
4488 
.5185 
5880 
6574 
7268 
7960 
8651 

799341 
800029 
0717 
1404 
2089 
2774 
3457 
4139 
4821 
5501 

N.  1   0   ll   |2|3|4j5|6|7|8|9| 

D. 

*^ 


A  TABLE  OF  LOGARITHMS  PROM  I  TO  10,000. 


11 


N. 

1   0 

1|2|3|-4|5|6|7|8I9|D.  1 

640 
641 
642 
643 
644 
645 
646 
647 
648 
649 
650 
651 
652 
653 
654 
655 
656 
657 
658 
659 
660 
661 
662 
663 
664 
665 
666 
667 
668 
669 
670 
671 
672 
073 
674 
675 
676 
677 
678 
679 
680 
681 
682 
683 
684 
685 
686 
687 
688 
689 
690 
691 
692 
693 
694 
695 
696 
697 
698 
699 

806180 
6858 
7535 
8211 
8886 
9560 

810233 
0904 
1575 
2245 

6248 
6926 
7603 
8279 
8953 
9627 
0300 
0971 
1642 
2312 
2980 
3648 
4314 
4980 
5644 
6308 
6970 
7631 
8292 
8951 
9610 
0267 
0924 
1579 
2233 
2887 
3539 
4191 
4841 
5491 
6140 
6787 
7434 
8080 
8724 
9368 
..11 
0653 
1294 
1934 
2573 
3211 
3848 
4484 
5120 
5754 
6387 
7020 
7652 
8282 

6316 
6994 
7670 
8346 
9021 
9694 
0367 
1039 
1709 
2379 

16384 
7061 
7738 
8414 
9088 
9762 
0434 
1106 
1776 
2445 

6451 
7129 
7806 
8481 
9156 
9829 
0501 
1173 
1843 
2512 

6519 
7197 
7873 
8549 
9223 
9896 
0569 
1240 
1910 
2579 
3247 
3914 
4581 
5246 
5910 
6573 
7235 
7896 
8556 
9215 

6587 
7264 
7941 
8616 
9290 
9964 
0636 
1307 
1977 
2646 
3314 
3981 
4647 
5312 
5976 
6639 
7301 
7962 
8622 
9281 

6655 
7332 

8008 
8684 
9358 
..31 
0703 
1374 
2044 
2713 
3381 
4048 
4714 
5378 
6042 
6705 
7367 
8028 
8688 
9346 
...4 
0661 
1317 
1972 
2626 
3279 
3930 
4581 
5231 
5880 

6723 
7400 
8076 
8751 
9425 
..98 
0770 
1441 
2111 
2780 
3448 
4114 
4780 
5445 
6109 
6771 
7433 
8094 
8754 
9412 
..70 
0727 
1382 
2037 
2691 
3344 
3996 
4646 
5296 
5945 
6593 
7240 
7886 
8531 
9175 
9818 
.460 
1102 
1742 
2381 

6790 
7467 
8143 
8818 
9492 
.165 
0837 
1508 
2178 
2847 
.3514 
4181 
4847 
5511 
6175 
6838 
7499 
8160 
8820 
9478 
.136 
0792 
1448 
2103 
2756 
3409 
4061 
4711 
5361 
6010 
6658 
7305 
7951 
8595 
9239 
9882 
.525 
1166 
1806 
2445 

68 
68 
68 
67 
67 
67 
67 
67 
67 
67 
67 
67 
67 
66 
66 
66 
66 
66 
66 
66 
66 
66 
66 
65 
65 
65 
65 
65 
65 
65 
65 
05 
65 
64 
64 
64 
64 
64 
64 
64 
64 
64 
64 
64 
63 
63 
63 
63 
63 
63 
63 
63 
63 
63 
63 
62 
62 
62 
62 
62 

81291J>1 
3581 
4248 
4913 
5578 
6241 
6904 
7565 
•8226 
8885 

3047 
3714 
4381 
5046 
5711 
6374 
7036 
7698 
8358 
9017 
9676 
0333 
0989 
1645 
2299 
2952 
3605 
4256 
4906 
5556 
6204 
6852 
7499 
8144 
8789 
9432 
..75 
0717 
1358 
1998 

3114 
3781 
4447 
5113 
5777 
6440 
7102 
7764 
8424 
9083 
9741 
0399 
1055 
1710 
2364 
3018 
3670 
4321 
4971 
5621 

3181 
3S48 
4514 
5179 
5843 
6506 
7169 
7830 
8490 
9149 

819544 
820201 
0858 
1514 
2168 
2822 
3474 
4126 
4776 
5426 

9807 
0464 
1120 
1775 
2430 
3083 
3735 
4386 
5036 
5686 
6334 
6981 
7628 
8273 
8918 
9561 
.204 
0845 
1486 
2126 
2764 
3402 
4039 
4675 
5310 
5944 
6577 
7210 
7841 
8471 

9873 
0530 
1186 
1841 
2495 
3148 
3800 
4451 
5101 
5751 
6399 
7046 
7692 
8338 
8982 
9625 
.268 
0909 
1550 
2189 

9939 
0595 
1251 
1906 
2560 
3213 
3865 
4516 
5166 
5815 
6464 
7111 
7757 
8402 
9046 
9690 
.332 
0973 
1614 
2253 
2892 
3530 
4166 
4802 
5437 
6071 
6704 
7336 
7967 
8597 
9227 
9855 
0482 
1109 
1735 
2360 
2983 
3606 
4229 
4850 

826075 
6723 
7369 
8015 
8660 
9304 
9947 

830589 
1230 
1870 

832509 
3147 
3784 
4421 
5056 
5691 
6324 
6957 
7588 
8219 

838849 
9478 

840106 
0733 
1359 
1985 
2609 
3233 
3855 
4477 

6269 
6917 
7563 
8209 
8853 
9497 
.139 
0781 
1422 
2062 
2700 
3338 
3975 
4611 
5247 
5881 
6514 
7146 
7778 
8408 

6528 
7175 
7821 
8467 
9111 
9754 
•396 
1037 
1678 
2317 
2956 
3593 
4230 
4866 
5500 
6134 
6767 
7399 
8030 
8660 

2637 
3275 
3912 
4548 
5183 
5817 
6451 
7083 
7715 
8345 

2828 
3466 
4103 
4739 
5373 
6007 
6641 
7273 
7904 
8534 
9164 
9792 
0420 
1046 
1672 
2297 
2921 
3544 
4166 
4788 

3020 
3657 
4294 
4929 
5564 
6197 
6830 
7402 
8093 
8723 
9352 
9981 
0608 
1234 
1860 
2484 
3108 
3731 
4353 
4974 

3083 
3721 
4357 
4993 
5627 
6261 
6894 
7525 
8156 
8786 

8912 
9541 
0169 
0796 
1422 
2047 
2672 
3295 
3918 
4539 

8975 
9604 
0232 
0859 
1485 
2110 
2734 
3357 
3980 
4601 

9038 
9667 
0294 
0921 
1547 
2172 
2796 
3420 
4042 
4664 

9101 
9729 
0357 
0984 
1610 
2235 
2859 
3482 
4104 
4726 

9289 
9918 
0545 
1172 
1797 
2422 
3046 
3669 
4291 
4912 

9415 
..43 
0671 
1297 
1922 
2547 
3170 
3793 
4415 
5036 

N.  1   0   I  1  1  2  1  3  1  4  1  5  1  6  1  7  1  8  1  9  1  D.  1 

12 


A  TABLE  OP  LOGARITHMS  FROM  1  TO  10,000 


N. 

1   0   1  1  1  2  1  3  1  4  1  5  1  6  1  7  1  8  1  9  |D.l 

700 

845098 

5160.  5222 

5284 

5346 1 5408 

5470  5532 

5594 

5656  62 1 

701 

6718 

5780 

5842 

5904 

5966 

6028 

6090 

6151 

6213 

6275 

62 

702 

6337 

6399 

6461 

6523 

6585 

6646 

6708 

6770 

6832 

6894 

62 

703 

6955 

7017 

7079 

7141 

7202 

7264 

7320 

7388 

7449 

7511 

62 

704 

7573 

7634 

7696 

7758 

7819 

7881 

7943 

8004 

8066 

8128 

62 

705 

8189 

8251 

8312 

8374 

8435 

8497 

8559 

8620 

8682 

8743 

62 

706 

8805 

8866 

8928 

8989 

9051 

9112 

9174 

9235 

9297 

9358 

61 

707 

9419 

9481 

9542 

9604 

9665 

9726 

9788 

9849 

9911 

9972 

61 

708 

850033 

0095 

0156 

0217 

0279 

0340 

0401 

0462 

0524 

0585 

61 

709 
710 

0646 

0707 
1320 

0769 
1381 

0830 

0891 

0952 
1564 

1014 
1625 

1075 

1136 

1747 

1197 
1809 

61 
61 

851258 

1442 

1503 

1686 

711 

1870 

1931 

1992 

2053 

2114 

2175 

2236 

2297 

2358 

2419 

61 

712 

2480 

2541 

2602 

2663 

2724 

2785 

2846 

2907 

2968 

3029 

61 

713 

3090 

3150 

3211 

3272 

3333 

3394 

3455 

3516 

3577 

3637 

61 

714 

3698 

3759 

3820 

3881 

3941 

4002 

4003 

4124 

4185 

4245 

61 

715 

4306 

4367 

4428 

4488 

4549 

4610 

4670 

4731 

4792 

4852 

61 

716 

4913 

4974 

5034 

5095 

5156 

5216 

5277 

5337 

5398 

5459 

61 

717 

5519 

5580 

5640 

5701 

5761 

5822 

5882 

5943 

6003 

6064 

61 

718 

6124 

6185 

6245 

6306 

6366 

6427 

6487 

6548 

6608 

6668 

60 

719 
720 

6729 
857332 

6789 
7393 

6850 

6910 
7513 

6970 

7574 

7031 
7634 

7091 

7152 

7212 

7272 
7875 

60 
60 

7453 

7694 

7755 

7815 

721 

7935 

7995 

8056 

8116 

8176 

8236 

8297 

8357 

8417 

8477 

60 

722 

8537 

8597 

8657 

8718 

8778 

8838 

8898 

8958 

9018 

9078 

60 

723 

9138 

9198 

9258 

9318 

9379 

9439 

9499 

9559 

9619 

9679 

60 

724 

9739 

9799 

9859 

9918 

9978 

..38 

..98 

.158 

.218 

.278 

60 

725 

860338 

0398 

0458 

0518 

0578 

0637 

0697 

0757 

0817 

0877 

60 

726 

0937 

0996 

1056 

1116 

1176 

1236 

1295 

1355 

1415 

1475 

60 

727 

1534 

1594 

1654 

1714 

1773 

1833 

1893 

1952 

2012 

2072 

60 

728 

2131 

2191 

2251 

2310 

2370 

2430 

2489 

2549 

2608 

2668 

60 

729 
730 

2728 

2787 
3382 

2847 

2906 

2966 

3025 

3085 

3144 
3739 

3204 

3263 

60 
59 

863323 

3442 

3501 

3561 

3620 

3080 

3799 

3858 

731 

3917 

3977 

4036 

4096 

4155 

4214 

4274 

4333 

4392 

4452 

59 

732 

4511 

4570 

4630 

4689 

4748 

4808 

4867 

4926 

4985 

5045 

59 

733 

5104 

5163 

5222 

5282 

5341 

5400 

5459 

5519 

5578 

5637 

59 

734 

5696 

5755 

5814 

5874 

5933 

5992 

6051 

6110 

6169 

6228 

59 

735 

6287 

6346 

6405 

6465 

6524 

6583 

6642 

6701 

6760 

6819 

59 

736 

6878 

6937 

6996 

7055 

7114 

7173 

7232 

7291 

7350 

7409 

59 

737 

7467 

7526 

7585 

7644 

7703 

7762 

7821 

7880 

7939 

7998 

59 

738 

8056 

8115 

8174 

8233 

8292 

8350 

8409 

8468 

8527 

8586 

59 

739 
740 

8644 

8703 
9290 

8762 

8821 

8879 
9466 

8938 
9525 

8997 
9584 

9056 

9114 

9173 

59 
59 

869232 

9349 

9408 

9642 

9701 

9760 

741 

9818 

9877 

9935 

9994 

..53 

.111 

.170 

.228 

.287 

.345 

59 

742 

870404 

0462 

0521 

0579 

0638 

0696 

0755 

0813 

0872 

0930 

68 

743 

0989 

1047 

1106 

1164 

1223 

1281 

1339 

1398 

1456 

1515 

58 

744 

.  1573 

1631 

1690 

1748 

1806 

1865 

1923 

1981 

2040 

2098 

68 

745 

2156 

2215 

2273 

2331 

2389 

2448 

2506 

2564 

2622 

2681 

68 

746 

2739 

2797 

2855 

2913 

2972 

3030 

3088 

3146 

3204 

3262 

68 

747 

3321 

3379 

3437 

3495 

3553 

3611 

3669 

3727 

3785 

3844 

68 

748 

3902 

3960 

4018 

4076 

4134 

4192 

4250 

4308 

4366 

4424 

58 

749 
750 

4482 

4540 

4598 

4656 

4714 
5293 

4772 
5351 

4830 

4888 

4945 

5003 

5582 

58 
68 

875061 

5119 

5177 

5235 

5409 

5466 

5524 

751 

5640 

5698 

5756 

5813 

5871 

5929 

5987 

6045 

6102 

6160 

68 

752 

6218 

6276 

6333 

6391 

6449 

6507 

6564 

6622 

6680 

6737 

68 

753 

6795 

6353 

6910 

6908 

7026 

7083 

7141 

7199 

7256 

7314 

68 

754 

7371 

7429 

7487 

7544 

7602 

7659 

7717 

7774 

7832 

7889 

68 

755 

7947 

8004 

8062 

8119 

8177 

8234 

8292 

8349 

8407 

8464 

67 

756 

8522 

8579 

8637 

8694 

8752 

8809 

8866 

8924 

8981 

9039 

67 

757 

9096 

9153 

9211 

9268 

9325 

9383 

9440 

9497 

9555 

9612 

67 

758 

9669 

9726 

9784 

9841 

9898 

9956 

..13 

..70 

.127 

.185 

67 

759 

880242 

0299 

0356 

0413 

0471 

0528 

0585 

0642 

0699 

0756 

67 

N. 

0   |ll2|3|4|5|6|7|8|9!D.  1 

A  TABLE  OF  LOGAEITHBIS  FROM  1 

TO  10,000 

13 

N. 

0   |l|2|3|4|5|6|7|8|9|D.  1 

760 

880814  0871.0928,0985 

1042  1099, 1156 

1213 

1271 

1328  57 1 

761 

1385 

1442 

1499 

1556 

1613 

1670 

1727 

1784 

1841 

1898 

57 

762 

1955 

2012 

2069 

2126 

2183 

2240 

2297 

2354 

2411 

2468 

67 

763 

2525 

2581 

2638 

2695 

2752 

2809 

2866 

2923 

2980 

3037 

57 

764 

3093 

3150 

3207 

3264 

3321 

3377 

3434 

3491 

3548 

3605 

57 

765 

3661 

3718 

3775 

3832 

3888 

3945 

4002 

4059 

4115 

4172 

57 

766 

4229 

4285 

4342 

4399 

4455 

4512 

4569 

4625 

4682  4739 

57 

767 

4795 

4852 

4909 

4965 

5022 

5078 

5135 

5192 

5248 

5305 

57 

768 

5361 

5418 

5474 

5531 

5587 

5644 

5700 

5757 

5813 

5870 

57 

769 

770 

5926 

5983 

6039 

6096 

6152 
6716 

6209 

6265 

6321 

6378 

6434 

56 
56 

886491 

6547 

6604 

6660 

6773 

6829 

6885 

6942 

6998 

771 

7054 

7111 

7167 

7223 

7280 

7336 

7392 

7449 

7505 

7561 

56 

772 

7617 

7674 

7730 

7786 

7842 

7898 

7955 

8011 

8067 

8123 

56 

773 

8179 

8236 

8292 

8348 

8404 

8460 

8516 

8573 

8629 

8685 

56 

774 

8741 

8797 

8853 

8909 

8965 

9021 

9077 

9134 

9190 

9246 

56 

775 

9302 

9358 

9414 

9470 

9526 

9582 

9638 

9694 

9750 

9806 

56 

776 

9862 

9918 

9974 

..30 

..86 

.141 

.197 

.253 

.309 

.365 

56 

777 

890421 

0477 

0533 

0589 

0645 

0700 

0756 

0812 

0868 

0924 

56 

T78 

0980 

1035 

1091 

1147 

1203 

1259 

1314 

1370 

1426 

1482 

56 

779 
780 

1537 

1593 
2150 

1049 
2206 

1705 

2262 

1760 
2317 

1816 
2373 

1872 

1928 

1983 

2039 
2595 

56 
56 

892095 

2429 

2484 

2540 

781 

2651 

2707 

2762 

2818 

2873 

2929 

2985 

3040 

3096 

3151 

56 

7*2 

3207 

3262 

3318 

3373 

3429 

3484 

3540 

3595 

3651 

3706 

56 

783 

3762 

3817 

3873 

3928 

3984 

4039 

4094 

4150 

4205 

4261 

55 

784 

4316 

4371 

4427 

4482 

4538 

4593 

4648 

4704 

4759 

4814 

55 

785 

4870 

4925 

4980 

5036 

5091 

5146 

5201 

5257 

5312 

5367 

55 

786 

5423 

5478 

5533 

5588 

5044 

5699 

5754 

5809 

5864 

5920 

55 

787 

5975 

6030 

6085 

6140 

6195 

6251 

6306 

6361 

6416 

6471 

65 

788 

6526 

6581 

6636 

0692 

6747 

6802 

6857 

6912 

6967 

7022 

65 

789 
790 

7077 

7132 

7187 
7737 

7242 
7792 

7297 

7352 

7407 

7462 

7517 

7572 

55 
55 

897627 

7682 

7847 

7902 

7957 

8012 

8067 

8122 

791 

8176 

8231 

8286 

8341 

8396 

8451 

8506 

8561 

8615 

8670 

65 

792 

8725 

8780 

8835 

8890 

8944 

8999 

9054 

9109 

9164 

9218 

66 

793 

9273 

9328 

9383 

9437 

9492 

9547 

9602 

9656 

9711 

9766 

55 

794 

9821 

9875 

9930 

9985 

..39 

..94 

.149 

.203 

.258 

.312 

56 

795 

900367 

0422 

0476 

0531 

0586 

0640 

0695 

0749 

0804 

0859 

65 

796 

0913 

0968 

1022 

1077 

1131 

1186 

1240 

1295 

1349 

1404 

65 

797 

1458 

1513 

1567 

1622 

1676 

1731 

1785 

1840 

1894 

1948 

54 

798 

2003 

2057 

2112 

2166 

2221 

2275 

2329 

2384 

2438 

2492 

54 

799 
800 

2547 

2601 

2655 
3199 

2710 
3253 

2764 
3307 

2818 

2873 

2927 
3470 

2981 
3524 

3036 

64 
54 

903090 

3144 

3361 

3416 

3578 

801 

3633 

3687 

3741 

3795 

3849 

3904 

3958 

4012 

4066 

4120 

64 

802 

4174 

4229 

4283 

4337 

4391 

4445 

4499 

4553 

4607 

4661 

54 

803 

4716 

4770 

4824 

4878 

4932 

4986 

5040 

5094 

5148 

5202 

54 

804 

5256 

5310 

5364 

6418 

5472 

5526 

5580 

5634 

5688 

5742 

54 

805 

5796 

5850 

5904 

5958 

6012 

6066 

6119 

6173 

6227 

6281 

54 

806 

6335 

6389 

6443 

6497 

6551 

6604 

66.58 

6712 

6766 

6820 

54 

807 

6874 

6927 

6981 

7035 

7089 

7143 

7196 

7250 

7304 

7358 

64 

808 

7411 

7465 

7519 

7573 

7626 

7680 

7734 

7787 

7841 

7895 

54 

809 
810 

7949 
908485 

8002 
8539 

8056 
8592 

8110 

8163 

8217 
8753 

8270 

8324 

8378 
8914 

8431 
8967 

54 
54 

8646 

8699 

8807 

8860 

811 

9021 

9074 

9128 

9181 

9235 

9289 

9342 

9396 

9449 

9503 

54 

812 

9556 

9610 

9663 

9716 

9770 

9823 

9877 

9930 

9984 

..37 

53 

813 

910091 

0144 

0197 

0251 

0304 

0358 

0411 

0464 

0518 

0571 

53 

814 

0624 

0678 

0731 

0784 

0838 

0891 

0944 

0998 

1051 

1104 

53 

815 

1158 

1211 

1264 

1317 

1371 

1424 

1477 

1530 

1584 

1637 

53 

816 

1690 

1743 

1797 

1850 

1903 

1956 

2009 

2063 

2116 

2169 

53 

817 

2222 

2275 

2328 

2381 

2435 

2488 

2541 

2594 

2647 

2700 

53 

818 

2753 

2806 

2859 

2913 

2966 

3019 

3072 

3125 

3178 

3231 

53 

819 

3284 

3337 

3390 

34431 3496 

3549 

3602 

3655 

3708 

3761 

63 

N. 

i   0 

1 

FT" 

I  3 

1  4 

5 

1  6 

1  7 

1  8 

1  9 

D.t 

Bb 


14 

A  TABLE  OP  LOGARITHMS  FROM  1  TO  10,000. 

N.  1   0   |i!2|3|4|5|6|7|8|9|D.  j 

820 
821 
822 
823 
824 
825 
826 
827 
828 
829 
830 
831 
832 
833 
834 
835 
836 
837 
838 
839 
840 
841 
842 
843 
844 
845 
846 
847 
848 
849 
850 
851 
852 
853 
854 
855 
856 
857 
858 
859 
860 
861 
862 
863 
864 
865 
866 
867 
868 
869 
870 
871 
872 
873 
874 
875 
876 
877 
878 
879 

913814 
4343 
4872 
5400 
5927 
6454 
6980 
7506 
8030 
8555 

3867 
4396 
4925 
5453 
5980 
6507 
7033 
7558 
8083 
8607 

3920 
4449 
4977 
5505 
6033 
6559 
7085 
7611 
&135 
8659 
9183 
9706 
0228 
0749 
1270 
1790 
2310 
2829 
3348 
3865 
4383 
4899 
5415 
5931 
6445 
6959 
7473 
7986 
8498 
9010 

3973 
4502 
503O 
5558 
6085 
6612 
7138 
7663 
8188 
8712 

4026 
4555 
.5083 
5611 
6138 
6664 
7190 
7716 
8240 
8764 
9287 
9810 
0332 
0853 
1374 
1894 
2414 
2933 
3451 
3969 

4079 
4608 
5136 
5664 
6191 
6717 
7243 
7768 
8293 
8816 

4132 
4660 
5189 
5716 
6243 
6770 
7295 
7820 
8345 
8869 

4184 
4713 
5241 
5769 
6296 
6822 
734« 
7873 
8397 
8921 
9444 
9967 
0489 
1010 
1530 
2050 
2570 
3089 
3607 
4124 

4237 
4766 
5294 
5822 
6349 
6875 
7400 
7925 
8450 
8973 
9496 
..19 
0541 
1062 
1582 
2102 
2622 
3140 
3658 
4176 

4290 
4819 
5347 
5875 
6401 
6927 
7453 
7978 
8502 
9026 
9549 
..71 
0593 
1114 
1634 
2154 
2674 
3192 
3710 
4228 

53 
53 
53 
63 
53 
53 
53 
52 
52 
52 
52 
52 
52 
52 
52 
62 
52 
52 
62 
52 
52 
52 
56 
51 
51 
51 
61 
51 
61 
51 
51 
51 
51 
51 
61 
51 
51 
61 
51 
51 
50 
60 
50 
50 
50 
50 
50 
50 
50 
50 
50 
50 
60 
60 
60 
50 
60 
49 
49 
49 

919078 
9601 

920123 
0645 
1166 
1686 
2206 
2725 
3244 
3762 

9130 
9653 
0176 
0697 
1218 
1738 
2258 
2777 
3296 
3814 
4331 
484S 
5354 
5879 
6394 
6908 
742Z 
7935 
8447 
8959 

9235 
9758 
0280 
0801 
1322 
1842 
2362 
2881 
3399 
3917 
4434 
4951 
5467 
5982 
6497 
7011 
7524 
8037 
8549 
9061 
9572 
..83 
0592 
1102 
1610 
2118 
2626 
3133 
3639 
4145 

9340 
9862 
0384 
090G 
1426 
1946 
2466 
2985 
3503 
4021 

9392 
9914 
0436 
0958 
1478 
1998 
2518 
3037 
3555 
4072 

924279 
4796 
5312 

5828 
6342 
6857 
7370 
7883 
8396 
8908 

4486 
.5003 
5518 
6034 
6548 
7062 
7576 
8088 
8601 
9112 

4538 
5054 
5570 
6085 
6600 
7114 
7627 
8140 
8652 
9163 

4589 
5106 
5621 
6137 
6651 
7165 
7678 
8191 
8703 
9215 
9725 
.236 
0745 
1254 
1763 
2271 
2778 
3285 
3791 
4296 

4641 
5157 
5673 
6188 
6702 
7216 
7730 
8242 
8754 
9266 

4693 
5209 
5725 
6240 
6754 
7268 
7781 
8293 
8805 
9317 

4744 
5261 
5776 
6291 
6805 
7319 
7832 
8345 
8857 
9368 
9879 
.389 
0898 
1407 
1915 
2423 
2930 
3437 
3943 
4448 
4953 
5457 
5960 
6463 
6966 
7468 
7969 
8470 
8970 
9469 
9968 
0467 
0964 
1462 
1958 
2455 
2950 
3445 
3939 
4433 

929419 
9930 

930440 
0949 
1458 
1966 
2474 
2981 
3487 
3993 

9470 
9981 
0491 
1000 
1509 
3017 
2524 
3031 
3538 
4044 

9521 
..32 
0542 
1051 
1560 
2068 
2575 
3082 
3589 
4094 
4599 
5104 
5608 
6111 
6614 
7117 
7618 
8119 
8620 
9120 
9619 
0118 
0616 
1114 
1611 
2107 
2603 
3099 
3593 
4088 

9623 
.134 
0643 
1153 
1661 
2169 
2677 
3183 
3690 
4195 
4700 
5205 
5709 
6212 
6715 
7217 
7718 
8219 
8720 
9220 
9719 
0218 
0716 
1213 
1710 
2207 
2702 
3198 
3692 
4186 

9674 
.185 
0694 
1204 
1712 
2220 
2727 
3234 
3740 
4246 
4751 
5255 
5759 
6262- 
6765 
7267 
7769 
8269 
8770 
9270 

9776 
.287 
0796 
1305 
1814 
2322 
2829 
3335 
3841 
4347 
4852 
5356 
5860 
6363 
6865 
7367 
7869 
8370 
8870 
9369 
9869 
0367 
0865 
1362 
1859 
2355 
2851 
3346 
3841 
4335 

9827 
.338 
0847 
1356 
1865 
2372 
2879 
3386 
3892 
4397 

934498 
6003 
5507 
6011 
6514 
7016 
7518 
8019 
8520 
9020 
939519 
940018 
0516 
1014 
1511 
2008 
2504 
3000 
3495 
3989 

4549 
5054 
5558 
6061 
6564 
7066 
7568 
8069 
8570 
9070 
9569 
0068 
0566 
1064 
1561 
2058 
2554 
3049 
3544 
4038 

4650 
5154 
5658 
0162 
6665 
7167 
7668 
8169 
8670 
9170 
9669 
0168 
0666 
1163 
1660 
2157 
2653 
3148 
3643 
4137 

4801 
5306 
5809 
6313 
6815 
7317 
7819 
8320 
8820 
9320 

9819 
0317 
0815 
1313 
1809 
2306 
2801 
3297 
3791 
4285 

4902 
5406 
5910 
6413 
6916 
7418 
7910 
8420 
8920 
9419 
9918 
0417 
0915 
1412 
1909 
2405 
2901 
3396 
3890 
4384 

9769 
0267 
0765 
1263 
1760 
2256 
2752 
3247 
3742 
4236 

N.  1   0   1  1  1  2  1  3  1  4  1  5  1  6  t  7  1  8  '  9  1  D.  1 

w: 


»: 


A  TABLE  Of  LOGARITHMS  FROM  1  TO  10,000. 


15 


F" 

0   lll2l3|4|5|6}7|8|9|D.  1 

880 

944483  45321 

4581  46311 

4680 

4729 

4779 

48281 48771 

4927  49 1 

881 

4976 

5025 

5074 

5124 

5173 

5222 

5272 

5321 

5370 

5419  49 1 

882 

6469 

5518 

5567 

5616 

5665 

5715 

5764 

5813 

5862 

5912 

49 

883 

5961 

6010 

6059 

6108 

6157 

6207 

6256 

6306 

6354 

6403 

49 

884 

6452 

6501 

6551 

6600 

6649 

6698 

6747 

6796 

6845 

6894 

49 

885 

6943 

6992 

7041 

7090 

7140 

7189 

7238 

7287 

7336 

7385 

49 

886 

7434 

7483 

7532 

7581 

7630 

7679 

7728 

7VVV 

7826 

7875 

49 

887 

7924 

7973 

8022 

8070 

8119 

8168 

8217 

8266 

8315 

8364 

49 

888 

8413 

8462 

8511 

8560 

8609 

8657 

8706 

8756 

8804 

8853 

49 

889 
890 

8902 

8951 

8999 

9048 

9097 

9146 
9634 

9195 
9683 

9244 
9731 

9292 
9780 

9341 

49 
49 

949390 

9439 

9488 

9536 

9585 

9829 

891 

9878 

9926 

9975 

..24 

..73 

.121 

.170 

.219 

.267 

.316 

49 

892 

950365 

0414 

0462 

0511 

0560 

0608 

0657 

0706 

0764 

0803 

49 

893 

0851 

0900 

0949 

0997 

1046 

1095 

1143 

1192 

1240 

12«9 

49 

894 

1338 

1386 

1435 

1483 

1532 

1580 

1629 

1677 

1726 

1776 

49 

8<*5 

1823 

1872 

1920 

1969 

2017 

2066 

2114 

2163 

2211 

2260 

48 

896 

2308 

2356 

2405 

2453 

2502 

2550 

2599 

2647 

2696 

2744 

48 

897 

2792 

2841 

2889 

2938 

2986 

3034 

3083 

3131 

3180 

3228 

48 

898 

3276 

3325 

3373 

3421 

3470 

3518 

3566 

3615 

3663 

3711 

48 

899 
900 

3760 

3808 
4291 

3856 

3905 
4387 

3953 

4001 

4049 
4532 

4098 
4580 

4146 
4628 

4194 
4677 

48 
48 

954243 

4339 

4435 

4484 

901 

4725 

4773 

4821 

4869 

4918 

4966 

6014 

5062 

5110 

.5168 

48 

902 

6207 

5255 

5303 

5351 

5399 

5447 

6495 

5543 

5592 

5640 

48 

903 

5688 

5736 

5784 

5832 

.5880 

5928 

.5976 

6024 

^073 

6120 

48 

904 

6168 

6216 

6265 

6313 

6361 

6409 

6457 

6605 

6553 

6601 

48 

905 

6649 

6697 

6745 

6793 

6840 

6888 

6936 

6984 

7032 

7^80 

48 

906 

7128 

7176 

7224 

7272 

7320 

7368 

7416 

7464 

751.2 

7559 

48 

907 

7607 

7655 

7703 

7751 

7799 

7847 

7894 

7942 

7990 

8038 

48 

908 

8086 

8134 

8181 

8229 

8277 

8325 

8373 

8421 

8468 

8516 

48 

909 
910 

8564 

8612 
9089 

8659 
9137 

8707 

8755 

8803 

8850 

8898 
9375 

8946 

8994 
9471 

48 
48 

959041 

9185 

9232 

9280 

9328 

9423 

911 

9518 

9566 

9614 

9661 

9709 

9757 

9804 

9852 

9900 

9947 

48 

912 

9995 

..42 

..90 

.138 

-185 

,233 

.280 

.328 

.376 

.423 

48 

913 

960471 

0518 

0566 

0613 

0661 

0709 

0756 

.0804 

0851 

0899 

48 

914 

0946 

0994 

1041 

1089 

1136 

1184 

1231 

1279 

1326 

1374 

47 

916 

1421 

1469 

1516 

1563 

1611 

1658 

1706 

17.53 

1801 

1848 

47 

916 

1895 

1943 

1990 

2038 

208512132 

2180 

2227 

2275 

2322 

47 

917 

2369 

2417 

2464 

2511 

2559 

2606 

2653 

2701 

2748 

2795 

47 

918 

2843 

2890 

2937 

2985 

3032 

3079 

3126 

3174 

3221 

3268 

47 

919 
920 

3316 

3363 

3410 

3882 

3457 

3504 
3977 

3552 
4024 

3599 

3646 
4118 

3693 
4165 

3741 
4212 

47 
47 

963788 

3835 

3929 

4071 

921 

4260 

4307 

4354 

4401 

4448 

4495 

4542 

4590 

4637 

4684 

47 

922 

4731 

4778 

4825 

4872 

4919 

4966 

5013 

5061 

5108 

5155 

47 

923 

5202 

5249 

5296 

5343 

5390 

5437 

5484 

5531 

5578 

5625 

47 

924 

5672 

5719 

5766 

5813 

5800 

5907 

5,954 

6001 

6048 

6095 

47 

925 

6142 

6189 

6236 

6283 

6329 

6376 

64?3 

6470 

6517 

6564 

47 

926 

6611 

6658 

6705 

6752 

6799 

6845 

6892 

6939 

6986 

7033 

47 

927 

7080 

7127 

7173 

7220 

7267 

7314 

7361 

7408 

7454 

7501 

47 

928 

7548 

7595 

7642 

7688 

7735 

7782 

7829 

7875 

7922 

7969 

47 

929 
930 

8016 

8062 

8109 

8576 

8156 
8623 

8203 
8670 

8249 
8716 

8296 

8343 

8810 

8390 
8856 

8436 

47 

47 

968483 

8530 

8763 

8903 

931 

8950 

8996 

9043 

9090 

9136 

9183 

9229 

9276 

9323 

9369 

47 

932 

9416 

9463 

9509 

9556 

9602 

9649 

9695 

9742 

9789 

9835 

47 

933 

9882 

9928 

9975 

..21 

..68 

.114 

.161 

.207 

.254 

.300 

47 

934 

970347 

0393 

0440 

0486 

0533 

0579 

0626 

0B72 

0719 

0765 

46 

935 

0812 

0858 

0904 

0951 

0997 

1044 

1090 

1137 

1183 

1229 

46 

936 

1276 

1322 

1369 

1415 

1461 

1508 

1554 

1601 

1647 

1693 

46 

937 

1740 

1786 

1832 

1879 

1925 

1971  2018  2064 

2110 

2157 

46 

938 

2203 

2249 

3295 

2342 

2388 

2434  2481  2527 

2573 

2619 

46 

939 

26661  2712'  2758>  2804'  2851'  2897'  2943'  2989'  3035'  3082'  46 

"nT 

1   0   lll2|3l4l5|6|7|8|9|D. 

16 

A  TABLE  OP  LOGARITHMS  FK03I  1 

TO  10,000 

. 

N. 

1   0   |l|2|3|4|5|6|7t8l9|D.  1 

940 

973128 

3174 

3220 

3266 

3313 

3359!  3405 

3451  3497 

3543 

46 

941 

3590 

3636 

3682 

3728 

3774 

3820 

3866 

3913 

3959 

4005 

46 

94a 

4051 

4097 

4143 

4189 

4235 

4281 

4327 

4374 

4420 

4466 

46 

943 

4512 

4558 

4604 

4650 

4696 

4742 

4788 

4834 

4880 

4926 

46 

944 

4972 

5018 

5064 

5110 

5156 

5202 

5248 

5294 

5340 

.5386 

46 

945 

5432 

5478 

6524 

5570 

5616 

5662 

5707 

5753 

5799 

5845 

46 

946 

5891 

5937 

5983 

6029 

6075 

6121 

6167 

6212 

62.58 

6.304 

46 

947 

6350 

6396 

6442 

6488 

6533 

6579 

6625 

6671 

6717 

6763 

46 

948 

6808 

6854 

6900 

6946 

6992 

7037 

7083 

71291 7175 

7220 

46 

949 
950 

7266 

7312 

7358 

7403 

7449 

7495 

7541 

7586 

7632 
8089 

7678 
81.35 

46 
46 

977724 

7769 

7815 

7861 

7906 

7:^52 

7998 

8043 

951 

8181 

8226 

8272 

8317 

8363 

8409 

8454 

8.500 

8546 

8591 

46 

952 

8637 

8683 

8728 

8774 

8819 

8865 

8911 

8956 

9002 

9047 

46 

953 

9093 

9138 

9184 

9230 

9275 

9.321 

9366 

9412 

9457 

9503 

46 

954 

9548 

9594 

9639 

9685 

9730 

9776 

9821 

9867 

9912 

99.58 

46 

955 

980003 

0049 

0094 

0140 

0185 

0231 

0276 

0322 

0367 

0412 

45 

956 

0458 

0503 

0549 

0594 

0640 

0685 

0730 

0776 

0821 

0867 

45 

957 

0912 

0957 

1003 

1048 

1093 

1139 

1184 

1229 

1275 

1320 

45 

958 

1366 

1411 

1456 

1501 

1547 

1.592 

1637 

1683 

1728 

1773 

45 

959 
960 

1819 

1864 
2316 

1909 
2362 

1954 

2000 

2045 

2090 

2135 

2181 

2226 

45 
45 

982271 

2407 

2452 

2497 

2543 

2588 

26.33 

2678 

961 

2723 

2769 

2814 

2859 

2904 

2949 

2994 

3040 

3085 

3130 

45 

962 

3175 

3220 

^265 

3310 

3356 

3401 

3446 

3491 

3536 

3581 

45 

963 

3626 

3671 

3716 

3762 

3807 

3852 

3897 

3942 

3987 

4032 

45 

964 

4077 

4122 

4167 

4212 

4257 

4302 

4347 

4392 

4437 

4-1:82 

45 

965 

4527 

4572 

4617 

4662 

4707 

4752 

4797 

4842 

4887 

4932 

45 

966 

4977 

5022 

5067 

5112 

5157 

5202 

5247 

5292 

53,37 

5382 

45 

967 

5426 

5471 

5516 

5561 

5606 

.5651 

5696 

5741 

5786 

5830 

45 

968 

5875 

5920 

5965 

6010 

6055 

6100 

6144 

6189 

6234 

6279 

45 

969 
970 

6324 

6369 

6413 

6458 

6.503 
6951 

6548 

6593 

6637 

7085 

6682 

6727 

45 

986772 

6817 

6861 

6906 

6996 

7040 

7130 

7175 

45 

971 

7219 

7264 

7309 

7353 

7398 

7443 

7488 

7.532 

7577 

7622 

45 

972 

7666 

7711 

7756 

7800 

7845 

7890 

7934 

7979 

8024 

8068 

45 

973 

8113 

8157 

8202 

8247 

8291 

8336 

8381 

8425 

8470 

8514 

45 

974 

8559 

8604 

8648 

8693 

8737 

8782 

8826 

8871 

8916 

8960 

45 

975 

9005 

9049 

9094 

9138 

9183 

9227 

9272 

9316 

9361 

9405 

45 

976 

9450 

9494 

9539 

9583 

9628 

9672 

9717 

9761 

9806 

9850 

44 

977 

9895 

9939 

9983 

..28 

..72 

.117 

.161 

.206 

.2.50 

.294 

44 

978 

990339 

0383 

0428 

0472 

0516 

0561 

0605 

0650 

0694 

0738 

44 

979 

0783 

0827 

0871 

0916 
13.59 

0960 

1004 

1049 

1093 

1137 
1,580 

1182 
1625 

44 
44 

991226 

1270 

1315 

1403 

1448 

1492 

1 .536 

981 

1669 

1713 

1758 

1802 

1846 

1890 

1935 

1979 

2023 

2067 

44 

982 

2111 

2156 

2200 

2244 

2288 

2.333 

2377 

2421 

2465 

2.509 

44 

983 

2554 

2598 

2642 

2686 

2730 

2774 

2819 

2863 

2907 

2951 

44 

984 

2995 

3039 

3083 

3127 

3172 

3216 

3260 

3304 

3348 

3392 

44 

985 

3436 

3480 

3524 

3568 

3613 

3657 

3701 

3745 

3789 

3833 

44 

986 

3877 

3921 

3965 

4009 

4053 

4097 

4141 

4185 

4229 

4273 

44 

987 

4317 

4361 

4405 

4449 

4493 

4.537 

4581 

4625 

4669 

4713 

44 

988 

4757 

4801 

4845 

4889 

4933 

4977 

.5021 

.5065 

5108 

5152 

44 

989 
990 

5196 
995635 

5240 
5679 

5284 

5328 
5767 

5372 

5416 

5460 

5504 
5942 

5547 

5591 

44 
44 

5723 

5811 

5854 

.5898 

5986 

6030 

991 

6074 

6117 

6161 

6205 

6249 

6293 

6337 

6380 

6424 

6468 

44 

992 

6512 

6555 

6599 

6643 

6687 

6731 

6774 

6818 

6862 

6906 

44 

993 

6949 

6993 

7037 

7080 

7124 

7168 

7212 

7255 

7299 

7343 

44 

994 

7386 

7430 

7474 

7517 

7.561 

7605 

7648 

7692 

7736 

7779 

44 

995 

7823 

7867 

7910 

7954 

7998 

8041 

8085 

8129 

8172 

8216 

44 

996 

8259 

8303 

8347 

8390 

8434 

8477 

8,521 

8564 

8608 

8652 

44 

997 

8695 

8739 

8782 

8826 

8869 

8913 

89.56 

9000 

9043 

9087 

44 

998 

9131 

9174 

9218 

926] 

9305 

9348 

9392 

9435 

9479 

9.522 

44 

999 

9565 

9609 

96521  9696 

97391  9783 

9826 

98701 9913 

9957 

43 

N. 

1   0   |l|2|3|4|6|6|7|8|9lD.  1 

A  TABLE 


OP 


LOGARITHMIC 
SINES   AND    TANGENTS, 


FOR   EVERY 


DEGREE    AND    MINUTE 


OP   THE    QUADRANT. 


N.B.  The  minutes  in  the  left-hand  column  of  each  page, 
increasing  downwards,  belong  to  the  degrees  at  the  top  ,•  and 
those  increasing  upwards,  in  the  right-hand  column,  belong  to 
the  degrees  below. 


l]b» 


18 

(0  L>egi 

*Ce.)   A  TABLE  OF   LOGARITHMIC 

17 

1   Sine 

1   D. 

1   Cosine   |  D. 

1   Tang. 

D. 

1   Cotang.  1   1 

=F 

0.000000 

10.000000 

0.000000 

liiiiniie. 

60 

1 

6.463726 

601717 

000000 

00 

6,463726 

501717 

13.5.36274 

69 

2 

764756 

293485 

000000 

00 

764756 

293483 

235244 

58 

3 

940847 

208231 

000000 

00 

940847 

208231 

059153 

57 

4 

7.065786 

161517 

000000 

00 

7.065786 

161517 

12.934214 

56 

5 

162696 

131968 

000000 

00 

162696 

131969 

837304 

55 

6 

241877 

111575 

9.999999 

01 

241878 

111578 

758122 

54 

7 

308824 

96653 

999999 

01 

308825 

996.53 

691175 

53 

8 

366816 

85254 

999999 

01 

366817 

85254 

633183 

52 

9 

417968 

76263 

9.99999 

01 

417970 

76263 

582030 

51 

10 
11 

463725 
7.505118 

68988 

999998 

01 
01 

463727 

68988 

636273 

50 
49 

62981 

9.999998 

7.505120 

62981 

12.494880 

12 

542906 

57936 

999997 

01 

542909 

57933 

457091 

48 

13 

577668 

53641 

999997 

01 

577672 

53642 

422328 

47 

14 

609853 

49938 

999996 

01 

609857 

49939 

390143 

46 

15 

639816 

46714 

999996 

01 

639820 

46715 

360180 

45 

16 

667845 

43881 

999995 

01 

667849 

43882 

332151 

44 

17 

694173 

41372 

999995 

01 

694179 

41373 

305821 

43 

18 

718997 

39135 

999994 

01 

719003 

39136 

280997 

42- 

19 

742477 

37127 

999993 

01 

742484 

37128 

257cl6 

41 

20 
21 

764754 

35315 

999993 
9.999992 

01 
01 

764761 

35136 

235239 

40 
39 

7.785943 

33672 

7.785951 

33673 

12.214049 

22 

806146 

32175 

999991 

01 

806155 

32176 

19.3845 

38 

23 

825451 

30805 

999990 

01 

825460 

30806 

174540 

37 

24 

843934 

29547 

999989 

02 

843944 

29549 

156056 

36 

25 

861662 

28388 

999988 

02 

861674 

28390 

138326 

35 

26 

878695 

27317 

999988 

02 

878708 

27318 

121292 

34 

27 

895085 

26323 

999987 

02 

895099 

26325 

104901 

33 

28 

910879 

25399 

999986 

02 

910894 

25401 

089106 

32 

29 

926119 

24538 

999985 

02 

926134 

24540 

073866 

31 

30 
31 

940842 

23733 
22980 

999983 
9.999982 

02 
02 

940858 
7.955100 

23735 

22981 

059142 

30 
29 

7.955082 

12.044900 

32 

968870 

22273 

999981 

02 

968889 

22275 

031111 

28 

33 

982233 

21608 

999980 

02 

982253 

21610 

017747 

27 

oM 

995198 

20981 

999979 

02 

995219 

20983 

004781 

26 

35 

8.007787 

20390 

999977 

02 

8.007809 

20392 

11.992191 

25 

36 

020021 

19831 

999976 

02 

020045 

19833 

979955 

24 

37 

031919 

19302 

999975 

02 

031945 

19305 

968055 

23 

38 

043501 

18801 

999973 

02 

043527 

18803 

956473 

22 

39 

054781 

18325 

999972 

02 

054809 

18327 

945191 

21 

40 
41 

065776 

17872 
17441 

999971 
9.999969 

02 

02 

065806 
8.076531 

17874 
17444 

934194 

20 
19 

8.076500 

11.923469 

42 

086965 

17031 

999968 

02 

086997 

17034 

913003 

18 

43 

097183 

16639 

999966 

02 

097217 

16042 

902783 

17 

44 

107167 

16265 

999964 

03 

107202 

16268 

892797 

16 

45 

116926 

15908 

999963 

03 

116963 

15910 

883037 

15 

46 

126471 

15566 

999961 

03 

126510 

15568 

873490 

14 

47 

135810 

15238 

999959 

03 

135851 

15241 

864149 

13 

48 

144953 

14924 

999958 

03 

144996 

14927 

855004 

12 

49 

153907 

14622 

999956 

03 

153952 

14627 

846048 

11 

5: 
51 

162681 

14333 

999954 

03 
03 

162727 

14336 

837273 
11.828672 

10 
9 

8.171280 

14054 

9.999952 

8.171328 

14057 

52 

179713 

13786 

999950 

03 

179703 

13790 

820237 

8 

53 

187985 

13529 

999948 

03 

188036 

13532 

811964 

7 

64 

196102 

13280 

999946 

03 

196156 

13284 

803844 

6 

55 

204070 

13041 

999944 

3 

204126 

13044 

795874 

5 

56 

211895 

12810 

999942 

4 

211953 

12814 

788047 

4 

57 

219581 

12587 

999940 

04 

219641 

12590 

780359 

3 

68 

227134 

12372 

999938 

04 

227195 

12376 

772805 

2 

59 

234557 

12164 

999936 

04 

234621 

12168 

765379 

1 

60 

241855 

11963 

999934  04 

241921 

11967 

758079 

0 

n 

Cosine  | 

Sine    1 

Colang. 

1 

Tang.   1  M.  | 

89  Degrees. 


SINES  AND  TANGE^'TS.   (1  DeglGG.] 

19 

M. 

Sine   )   D. 

Cosine   1  D. 

Tang.   1   D. 

Cotang.   1 

"0" 

8,241855 

11963 

9.999934 

04 

8.241921 

11967 

11.758079  60 

1 

249033 

11768 

999932 

04 

249102 

11772 

750898  59 

2 

256094 

11580 

999929 

04 

256165 

11584 

743835  58 

3 

263042 

11398 

999927 

04 

263115 

11402 

736885  57 

4 

269881 

11221 

999925 

04 

269956 

11225 

7300441  56 

5 

276614 

11050 

999922 

04 

276691 

11054 

723309 

55 

% 

283243 

10883 

999920 

04 

283323 

10887 

716677 

54 

7 

289773 

10721 

999918 

04 

28985G 

10726 

710144 

53 

8 

296207 

10565 

999915 

04 

296292 

10570 

703708 

52 

9 

302546 

10413 

999913 

04 

302634 

10418 

697366 

51 

10 
11 

308794 

10266 
10122 

999910 
9.999907 

04 
04 

308884 

10270 

691116 

50 
49 

8,314954 

8.315046 

10126 

11.684954 

12 

321027 

9982 

999905 

04 

321122 

9987 

678878 

48 

13 

327016 

9847 

999902 

04 

327114 

9851 

672886 

47 

14 

332924 

9714 

999899 

05 

333025 

9719 

666975 

46 

15 

338753 

9586 

999897 

05 

338S56J  9590 

661144 

45 

16 

344504 

9460 

999894 

05 

3446] 0|  9465 

655390 

44 

17 

350181 

9338 

999891 

05 

350289 

9343 

6,49711 

43 

18 

355783 

9219 

999888 

05 

355895 

9224 

644105 

42 

19 

'  361315 

9103 

999885 

05 

361430 

9108 

638570 

41 

20 
21 

366777 

8990 

999882 
9.999879 

05 
05 

366895 

8995 

633105 
11.627708 

40 
39 

8,372171 

8880 

8.372292 

8885 

22 

377499 

8772 

999876 

05 

377622 

8777 

622378 

38 

23 

382762 

8667 

999873 

05 

3828891  8672 

617111 

37 

24 

387962 

8564 

999870 

05 

388092 

8570 

611908 

36 

25 

393101 

8464 

999867 

05 

393234 

8470 

608766 

35 

26 

398179 

8366 

999864 

05 

398315 

8371 

601685 

34 

27 

403199 

8271 

999861 

05 

403338 

8276 

596662 

33 

^8 

408161 

8177 

999858 

05 

408304 

8182 

591696 

32 

^9 

413068 

8086 

999854 

05 

413213 

8091 

586787 

31 

30 
31 

417919 
8.422717 

7996 

999851 
9.999848 

06 
06 

418068 
8.422869 

8002 

581932 
11.577131 

30 
29 

7909 

7914 

32 

427462 

7823 

999844 

06 

427618 

7830 

572382 

28 

33 

432156 

7740 

999841 

06 

432315 

7745 

567685 

27 

34 

486800 

7657 

999838 

06 

436962 

7663 

563038 

26 

35 

441394 

7577 

999834 

06 

441560 

7583 

558440 

25 

36 

4;4594i 

7499 

999831 

06 

446110 

7505 

553890 

24 

37 

450440 

7422 

999827 

06 

450613 

7428 

549387 

23 

38 

454893 

7346 

999823 

06 

455070 

7352 

544930 

22 

39 

459301 

7273 

999820 

06 

459481 

7279 

540519 

21 

40 
41 

463665 
8.467985 

7200 

999816 
9.999812 

06 
06 

463849 
8.468172 

7206 

536151 

20 
19 

7129 

7135 

11.531828 

42 

472263 

7060 

999809 

06 

472454 

7066 

527546 

18 

43 

476498 

6991 

999805 

06 

476693 

6998 

523307 

17 

44 

480693 

6924 

99980] 

06 

480892 

6931 

5191.08 

16 

45 

484848 

6859 

999797 

07 

485050 

6865 

514950 

15 

46 

488963 

6794 

999793 

07 

489170 

6801 

510830 

14 

47 

493040 

6731 

999790 

07 

493250 

6738 

506750 

13 

48 

497078 

6669 

999788 

07 

497293 

6676 

502707 

12 

49 

501080 

6608 

999782 

07 

501298 

6615 

498702 

11 

50 
51 

505045 

6548 

999778 

07 
07 

505267 
8.509200 

6555 
6496 

494733 

10 
9 

8.508974 

6489 

9.999774 

11.490800 

52 

512867 

6431 

999769 

07 

513098 

6439 

486902 

8 

53 

516726 

6375 

999765 

07 

516961 

6382 

483039 

7 

54 

520551 

6319 

999761 

07 

5207P9 

6326 

479210 

6 

55 

524343 

6264 

999757 

07 

5245fc>  5 

C272 

475414 

5 

56 

528102 

6211 

999753 

07 

528349 

6218 

471651 

4 

67 

631828 

6158 

999748 

07 

532080 

6165 

467920 

3 

58 

535523 

6106 

999744 

07 

635779 

6113 

464221 

2 

59 

539186 

6055 

999740 

07 

539447 

6062 

460553 

1 

60 

542819 

6004 

999735 

07 

543084  6012 

456916 

0 

B 

Cosine  j 

Sine   j 

Cotang.  1 

Tang.    jAf.j 

rib  Degrees 


so 

(2  Degrees.)  a 

TABLE  OF  LOGARITHMIC 

T 

Sine   j   D. 

(Cosine   j  D. 

T;uiii. 

D. 

Cotanc.   1 

0 

8.542819 

6004 

9.999735 

07 

8.. 543084 

6012 

1 1.456916  GO 

1 

546422 

5965 

999731 

07 

546691 

5962 

453309 

59 

2 

549995 

6906 

999726 

07 

.550268 

6914 

449732 

58 

3 

553539 

6858 

999722 

08 

553817 

5866 

446183 

57 

4 

557054 

5811 

999717 

08 

557336 

5819 

442664 

56 

5 

560540 

6766 

999713 

08 

560828 

6773 

439172 

.55 

6 

563999 

5719 

999708 

08 

564291 

5727 

436709 

54 

7 

567431 

5674 

999704 

08 

567727 

5682 

432273 

53 

8 

570836 

5630 

999699  08 

571137 

56.38 

4288631  52  | 

9 

574214 

5587 

999694 

08 

574520 

5595 

426480 

51 

10 
11 

577566 
8.580892 

5544 
5602 

999689 
9.999086 

08 
08 

577877 

5552 

422123 

50 
49 

8.581208 

5510 

11.418792 

12 

584193 

5460 

999680 

08 

584514 

5468 

416486 

48 

13 

587469 

5419 

999675 

08 

587795 

5427 

412205 

47 

14 

690721 

.5379 

999670 

08 

591061 

5387 

408949 

46 

15 

593948 

5339 

999665 

08 

594283 

5347 

405717 

45 

16 

597152 

5300 

999660 

08 

697492 

6308 

402508 

44 

17 

600332 

5261 

999656 

08 

600677 

6270 

399323 

43 

18 

603489 

5223 

9996.50 

08 

603839 

5232 

396161 

42 

19 

606623 

6186 

999645 

09 

606978 

5194 

393022 

41 

20 

31 

609734 

5149 

999640 
9.999035 

09 
09 

610094 
8.613189 

51.58 

389906 

40 
39 

8.612823 

•5112 

5121 

11.386811 

22 

615891 

5076 

999629 

09 

616262 

5085 

383738 

38 

23 

618937 

5041 

999624 

09 

619313 

5050 

380687 

37 

24 

621962 

5006 

999619 

09 

622343 

.5016 

377657 

36 

25 

624965 

4972 

999614 

09 

626352 

4981 

374648 

35 

26 

627948 

4938 

999608 

09 

628340 

4947 

371660 

34 

27 

630911 

4904 

999603 

09 

631308 

4913 

368692 

33 

28 

633854 

4871 

999597 

09 

634256 

4880 

365744 

32 

29 

636776 

4839 

999592 

09 

637184 

4848 

362816 

31 

30 
31 

639680 
8.642563 

4806 

999586 
9.999681 

09 
09 

640093 

4816 

4784 

359907 

30 

4776 

8.642982 

11. 3570181  29  1 

32 

645428 

4743 

999576 

09 

645863 

4753 

364147 

28 

33 

648274 

4712 

999570 

09 

648704 

4722 

351290 

27 

34 

651102 

4682 

999664 

09 

651537 

4691 

348463 

26 

35 

653911 

4652 

999558 

10 

654352 

4661 

345648 

25 

36 

656702 

4622 

999563 

10 

067149 

4631 

342851 

24 

37 

659475 

4592 

999547 

10 

659928 

4602 

340072 

23 

38 

662230 

4563 

999541 

10 

662689 

4573 

837311 

22 

39 

664968 

4535 

999536 

10 

665433 

4544 

334567 

21 

40 

667689 

4606 

999629 

10 

668160 

4526 

331840 

20 

41 

8.670393 

4479 

9.999524 

10 

8.670870 

4488 

11.329130 

19 

42 

673080 

4451 

999518 

10 

673563 

4461 

326437 

18 

43 

675761 

4424 

999512 

10 

676239 

4434 

323761 

17 

44 

678405 

4397 

999506 

10 

678900 

4417 

321100 

16 

45 

681043 

4370 

999500 

10 

681544 

4380 

318456 

15 

46 

683666 

4344 

999493 

10 

684172 

4364 

316828 

14 

47 

686272 

4318 

999487 

10 

686784 

4328 

313216 

13 

48 

688803 

4292 

999481 

10 

689381 

4303 

310619 

12 

49 

691438 

4267 

999475 

10 

69-1963 

4277 

308037 

11 

50 
51 

693993 
8.696543 

4242 

9994G9 
9.990463 

10 

11 

694529 
8.697081 

4252 

305471 
11.302919 

10 
9 

4217 

4228 

52 

699073 

4192 

999456 

11 

699617 

4203 

300383 

8 

53 

701689 

4168 

999450 

11 

702139 

4179 

297861 

7 

54 

704090 

4144 

999443 

11 

704646 

4155 

295354 

6 

55 

706577 

4121 

999437 

11 

707140 

4132 

292860 

5 

56 

709049 

4097 

999431 

11 

709618 

4108 

29i>382 

4 

57 

711507 

4074 

999424 

11 

712083 

4085 

287917 

3 

58 

713962 

4051 

999418 

11 

714534 

4062 

285466 

2 

59 

716383 

4029 

999411 

11 

716972 

4040 

283028 

1 

60 

718800 

4006 

999404 

11 

71939G 

4017 

280604'  01 

L 

Cosine  | 

Sine   1 

(Jotaiic. 

1 

Tang.    |m7| 

87  Degrees. 


»♦ 


SINES  AND  TANGENTS.   (3  DcgrceS.) 

21 

M. 

Sine 

D. 

Cosine  |  D. 

'J'ans;. 

1   D. 

Cotanir.   |   | 

0 

8.718800 

4006 

9.999404  11 

8.719396 

4017 

11.280604 

60 

1 

721204 

3984 

999398  11 

721806 

3995 

278194 

59 

2 

723595 

3962 

999391 

11 

724204 

3974 

275796 

58 

3 

725972 

3941 

99^384 

11 

726588 

3952 

273412 

57 

4 

728337 

3919 

999378 

11 

728959 

3930 

271041 

56 

5 

730688 

3898 

999371 

11 

731317 

3909 

268683 

55 

6 

733027 

3877 

999364 

12 

733663 

3889 

266337 

54 

7 

735354 

3857 

999357 

12 

735996 

3868 

264004 

53 

8 

737667 

3836 

999350 

12 

738317 

3848 

261683 

52 

9 

739969 

3816 

999343 

12 

740626 

3827 

259374 

51 

10 
11 

742259 
8.744536 

3796 

999336 

12 
12 

742922 

3807 

257078 

50 
49 

3776 

9.999329 

8.745207 

3787 

11.254793 

12 

746802 

3756 

999322 

12 

747479 

3768 

252521 

48 

13 

749055 

3737 

999315 

12 

749740 

3749 

250260 

47 

14 

751297 

3717 

999308 

12 

751989 

3729 

248011 

46 

15 

753528 

3698 

999301 

12 

754227 

3710 

245773 

45 

16 

755747 

3679 

999294 

12 

756453 

3692 

243547 

44 

17 

757955 

3661 

999286 

12 

758668 

3673 

241332 

43 

18 

760151 

3642 

999279 

12 

760872 

3655 

239128 

42 

19 

762337 

3624 

999272 

12 

763065 

3636 

236935 

41 

20 
21 

764511 
8.766675 

3606 

999265 
9.999257 

12 
12 

765246 

3618 

234754 

40 
39 

3588 

8.767417 

3600 

11.232583 

22 

768828 

3570 

9992.50 

13 

769578 

3583 

230422 

38 

23 

770970 

3553 

999242 

13 

771727 

3565 

228273 

37 

24 

773101 

3.535 

999235 

13 

773866 

3548 

226134 

36 

25 

775223 

3518 

999227 

13 

775995 

3531 

224005 

35 

26 

777333 

3501 

999220 

13 

778114 

3514 

221886 

34 

27 

779434 

3484 

999212 

13 

780222 

3497 

219778 

33 

28 

781524 

3467 

999205 

13 

782320 

3480 

217680 

32 

29 

783605 

3451 

999197 

13 

784408 

3464 

215592 

31 

30 
31 

785675 

.3431 

999189 

13 
13 

786486 

3447 

213514 

30 
29 

8.787736 

3418 

9.999181 

8.788554 

3431 

11.211446 

32 

789787 

3402 

999174 

13 

790613 

3414 

209387 

28 

33 

791828 

3386 

999166 

13 

792662 

3399 

207338 

27 

34 

793859 

3370 

999158 

13 

794701 

3383 

205299 

26 

35 

795881 

3354 

999150 

13 

796731 

3368 

203269 

25 

36 

797894 

3339 

999142 

13 

798752 

3352 

201248 

24 

37 

799897 

3323 

999134 

13 

800763 

3337 

199237 

23 

38 

801892 

3308 

999126 

13 

802765 

3322 

197235 

22 

39 
40 

41 

S03876 
805852 

3293 
3278 

999118 

13 

804758 

3307 
3292 

195242 
193258 

21 
20 

19 

999110 

13 
13 

806742 

8.807819 

3263 

9.999102 

8.808717 

3278 

11.191283 

42 

809777 

3249 

999094 

14 

810683 

3262 

189317 

18 

43 

811726 

3234 

999086 

14 

812641 

3248 

187359 

17 

44 

813667 

3219 

999077 

14 

814589 

3233 

185411 

16 

45 

815599 

3205 

999069 

14 

816529 

3219 

183471 

15 

46 

817522 

3191 

999061 

14 

818461 

3205 

181539 

14 

47 

819436 

3177 

999053 

14 

820384 

3191 

179616 

13 

48 

821343 

3163 

999044 

14 

822298 

3177 

177702 

12 

49 

823240 

3149 

999036 

14 

824205 

3163 

175795 

11 

50 
51 

825130 
8.827011 

3135 

999027 

14 

14 

826103 

3150 
3136 

173897 

10 
9 

3122 

9.999019 

8.827992 

11,172008 

52 

828884 

3108 

999010 

14 

829874 

3123 

170126 

8 

53 

830749 

3095 

999002 

14 

831748 

3110 

168252 

7 

54 

832607 

3082 

998993 

14 

833613 

3096 

166387 

6 

55 

834456 

3069 

998984 

14 

835471 

3083 

164529 

5 

56 

836297 

3056 

998976 

14 

837321 

3070 

162679 

4 

57 

838130 

3043 

998967 

15 

839163 

3057 

160837 

3 

68 

839956 

3030 

998958 

15 

840998 

3045 

159002 

2 

59 

841774 

3017 

998950 

15 

842825 

3032 

167175  1 

60 

843585 

3000 

998941 

15 

8446441 

3019 

155356  0 

U 

Cosine 

Sine   1 

Cotang.  1 

1 

Tang.   1 M. 

86  Degrees. 


>¥•' 


22 

(4  Degrees.)  a 

TABLE  OF  LOGARITHMIC 

T| 

Sine 

D.   1 

Cosine   |  D.  | 

Tang.   1   D.   | 

Cotang.  1    1 

0 

8.843585 

3005 

9.998941 

15 

8.844644 

3019  11.1553661 

60 

1 

845387 

2992 

998932 

16 

846465 

3007 

153645 

69 

2 

847183 

2980 

998923 

15 

848260 

2996 

151740 

68 

3 

848971 

2P67 

998914 

15 

850067 

2982 

149943 

57 

4 

850751 

2956 

998906 

15 

851846 

2970 

148154 

56 

5 

862525 

2943 

998896 

16 

853628 

2968 

146372 

55 

6 

854291 

2931 

998887 

15 

866403 

2946 

144597 

54 

7 

856049 

2919 

998878 

15 

867171 

2935 

142829 

.53 

8 

857801 

2907 

998869 

15 

868932 

2923 

141068  .52  1 

9 

859546 

2896 

998860 

15 

860686 

2911 

139314 

51 

10 
11 

861283 
8.863014 

288 1 

998861 
9.998841 

16 
16 

862433 

2900 

2888 

137567 
11.135827 

60 
49 

2873 

8.864173 

12 

864738 

2861 

998832 

15 

866906 

2877 

134094 

48 

13 

866456 

2860 

998823 

16 

867632 

2866 

132368 

47 

14 

868165 

2839 

998813 

16 

869351 

2854 

130649 

46 

15 

869868 

2828 

998804 

16 

871064 

2843 

128936 

45 

16 

871565 

2817 

998796 

16 

872770 

2832 

127230 

44 

17 

873255 

2806 

998785 

16 

874469 

2821 

125531 

43 

18 

874938 

2795 

998776 

16 

876162 

2811 

123838 

42 

19 

876615 

2786 

998766 

16 

877849 

2800 

122161 

41 

20 
21 

878285 

2773 

998767 

16 
16 

879629 

2789 

120471 

40 
39 

8.879949 

2763 

9.998747 

8.881202 

2779 

11.118798 

22 

881607 

2762 

998738 

16 

882869 

2768 

117131 

38 

23 

883258 

2742 

998728 

16 

884630 

2758 

115470 

37 

24 

884903 

2731 

998718 

16 

886186 

2747 

113815 

36 

25 

886542 

2721 

998708 

16 

887833 

2737 

112167 

35 

26 

888174 

2711 

998699 

16 

889476 

2727 

110624 

34 

27 

889801 

2700 

998089 

16 

891112 

2717 

108888 

33 

28 

891421 

2690 

998679 

16 

892742 

2707 

107258 

32 

29 

893035 

2680 

998669 

17 

894366 

2697 

105634 

31 

30 
31 

894643 

2670 
2660 

998659 

17 
17 

895984 

2687 

104016 

30 
29 

8.896246 

9.998649 

8.897696 

2677 

11.102404 

32 

897842 

2661 

998639 

17 

899203 

2667 

100797 

28 

33 

899432 

2641 

998629 

17 

900803 

2658 

099197 

27 

34 

901017 

2631 

998619 

17 

902398 

2648 

097692 

26 

35 

902696 

2622 

998609 

17 

903987 

2638 

096013 

25 

36 

904169 

2612 

998599 

17 

905570 

2629 

094430 

24 

37 

905736 

2603 

998589 

17 

907147 

2620 

092863 

23 

38 

907297 

2593 

998578 

17 

908719 

2610 

091281 

22 

39 
40 
41 

908863 
910404 

2584 

998668 

17 

910286 

911846 

8.913401 

2601 
2592 

089716 
088164 

21 
20 

T9 

2575 

998558 

IT 
17 

8.911949 

2666 

9.998548 

2683 

11.086699 

42 

913488 

2556 

998537 

17 

914961 

2674 

085049 

18 

43 

915022 

2547 

998527 

17 

916495 

2665 

083.506 

17 

44 

916660 

2538 

998516 

18 

918034 

2566 

081966 

16 

45 

918073 

2629 

998506 

18 

919668 

2547 

080432 

15 

46 

919691 

2620 

998496 

18 

921096 

2638 

078904 

14 

47 

921103 

2612 

998486 

18 

922619 

2630 

077381 

13 

48 

922610 

2603 

998474 

18 

924136 

2621 

075864 

12 

49 

924112 

2494 

998464 

18 

925649 

2512 

074351 

11 

50 
61 

926609 
8.927100 

2486 
24*77 

998453 

18 
18 

927166 
8.928668 

2603 

072844 

10 
9 

9.998442 

2495 

11.071342 

52 

928587 

2469 

998431 

18 

930166 

2486 

069845 

8 

53 

930068 

2460 

998421 

18 

931647 

2478 

068363 

7 

54 

931544 

2462 

998410 

18 

933134 

2470 

066866 

6 

55 

933015 

2443 

998399 

18 

934616 

2461 

065384 

5 

56 

934481 

2436 

998388 

18 

936093 

2453 

U63907 

4 

57 

935942 

2427 

998377 

18 

937566 

2445 

062435 

3 

58 

937398 

2419 

998366 

18 

939032 

2437 

060968 

2 

59 

93885C 

2411 

998355 

18 

940494 

2430 

069506 

1 

60 

940296 

2403 

998344 

18 

941962 

2421 

068048 

0 

1  Cosine 

1 

1   Sine   1 

1  Cotang.  1 

1   Tang.   |M.| 

85  Degrees. 


SINES  AND  TANGENTS.  (5  Degrees.) 

23 

M 

Sine 

D. 

Cosine   1  D. 

1   Tang. 

1   D. 

1  Ootang;.   |   | 

0 

8.940296 

2403 

9.998344 

19 

8.941952 

2421 

11.058048 

60 

1 

941738 

2394 

998333 

19 

943404 

2413 

056596 

59 

2 

943174 

2387 

998322 

19 

944S52 

2405 

055148 

58 

3 

944606 

2379 

998311 

19 

946295 

2397 

053705 

57 

4 

946034 

2371 

998300 

19 

947734 

2390 

052266 

56 

?i 

947456 

2363 

998289 

19 

949168 

2382 

050832 

55 

h 

948874 

2355 

998277 

19 

950597 

2374 

049403 

54 

V 

950287 

2348 

99826G 

19 

952021 

2366 

047979 

53 

p 

951696 

2340 

998255 

19 

953441 

2360 

046559 

52 

c 

953100 

2332 

998243 

19 

954856 

2351 

045144 

51 

10 
ll' 

954499 
8.955894 

2325 

998232 
9.998220 

19 
19 

956267 
8.957674 

2344 

043733 

50 
49 

2317 

2337 

11.042326 

15 

957284 

2310 

998209 

19 

959075 

2329 

040925 

48 

U- 

958670 

2302 

998197 

19 

960473 

2323 

039527 

47 

14 

960052 

2295 

998186 

19 

961866 

2314 

038134 

46 

If' 

961429 

2288 

998174 

19 

963255 

2307 

036745 

45 

If. 

962801 

2280 

998163 

19 

964639 

2300 

035361 

44 

ir 

964170 

2273 

998151 

19 

966019 

2293 

033981 

43 

18 

965534 

2266 

998139 

20 

967394 

2286 

032606 

42 

19 

966893 

2259 

998128 

20 

968766 

2279 

0312.34 

41 

20 
21 

968249 
8.969600 

2252 

998116 
9.998104 

20 
20 

9701.33 
8.971496 

2271 

029867 

40 
39 

2244 

2265 

11.028504 

22 

970947 

2238 

998092 

20 

972855 

2257 

027145 

38 

23 

972289 

2231 

998080 

20 

974209 

2251 

025791 

37 

24 

973628 

2224 

998068 

20 

975560 

2244 

024440 

36 

25 

974962 

2217 

998056 

20 

976906 

2237 

023094 

35 

26 

976293 

2210 

998044 

20 

978248 

2230 

021752 

34 

27 

977619 

2203 

998032 

20 

979586 

2223 

020414 

33 

28 

978941 

2197 

998020 

20 

980921 

2217 

019079 

32 

29 

980259 

2190 

998008 

20 

982251 

2210 

017749 

31 

30 
31 

981573 

8.982883 

2183 

997996 
9.997984 

20 
20 

983577 

,2204 

016423 

30 
29 

2177 

8.984899 

2197 

11.015101 

32 

984189 

2170 

997972 

20 

986217 

2191 

013783 

28 

33 

985491 

2163 

997959 

20 

987532 

2184 

012468 

27 

34 

986789 

2157 

997947 

20 

988842 

2178 

011158 

26 

35 

988083 

2150 

997935 

21 

990149 

2171 

009851 

25 

36 

989374 

2144 

997922 

21 

991451 

2165 

008549 

24 

37 

990660 

2138 

997910 

21 

992750 

21.58 

007250 

23 

38 

991943 

2131 

997897 

21 

994045 

2152 

005955 

22 

39 

993222 

2125 

997885 

21 

995337 

2146 

004663 

21 

40 
41 

994497 
8.995768 

2119 
2112 

997872 
9.997860 

21 
21 

996624 

2140 

003376 

20 
19 

8.997908 

21,34 

11.002092 

42 

997036 

2106 

997847 

21 

999188 

2127 

000812 

18 

43 

998299 

2100 

997835 

21 

9.000465 

2121 

10.999535 

17 

44 

999560 

2094 

997822 

21 

001738 

2115 

998262 

16 

45 

9.000816 

2087 

997809 

21 

003007 

2109 

996993 

15 

46 

002069 

2082 

997797 

21 

004272 

2103 

995728 

14 

47 

003318 

2076 

997781 

21 

005534 

2097 

994466 

13 

48 

004563 

2070 

997771 

21 

006792 

2091 

993208 

12 

49 

005805 

2064 

997758 

21 

008047 

2085 

991953 

11 

50 
51 

007044 
9.008278 

2058 

997745 

21 
21 

009298 
9.010546 

2080 

990702 
10.989454 

0 
9 

2052 

9.997732 

2074 

52 

009510 

2046 

997719 

21 

011790 

2068 

988210 

8 

53 

010737 

2040 

997706 

21 

013031 

2062 

986969 

7 

54 

011962 

2034 

997693 

22 

014268 

2056 

985732 

6 

55 

013182 

2029 

997680 

22 

015502 

2051 

984498 

5 

56 

014400 

2023 

997667 

22 

016V32 

2045 

983268 

4 

57 

015613 

2017 

997654 

22 

017959 

2040 

982041 

3 

58 

016824 

2012 

997641 

22 

019183 

2033 

980817 

2 

59 

018031 

2006 

997628 

22 

020403 

2028 

979597 

1 

60 

019235 

2000 

997614 

22 

021620 

2023 

978380 

0 

^ 

Cosine  | 

1 

Sine   1   1 

Cotang.  1 

1 

Tang.   |M.  I 

84  Degrees. 


24 

(6  Degrees.)   a 

FABLE  OF  LOGABITHMIC 

;^ 

Sine 

D. 

Cosine   |  1). 

Tang. 

D.   1 

Cotang.   1   1 

0 

9.019235 

200p 
1995 

9.997614 

22 

9.021620 

2023 

10.978380 

60 

1 

020435 

997601 

22 

022834 

2017 

977166 

59 

2 

021632 

1989 

997588 

22 

024044 

2011 

975956 

58 

3 

022825 

1984 

997574 

22 

025251 

2006 

974749 

57 

4 

024016 

1978 

997561 

22 

026455 

2000 

973545 

56 

5 

025203 

1973 

997547 

22 

027655 

1995 

972345 

55 

6 

026386 

1967 

997534 

23 

028852 

1990 

971148 

54 

7 

027567 

1962 

997520 

23 

030046 

1985 

969954 

53 

8 

028744 

1957 

997507 

23 

031237 

1979 

968763 

52 

9 

029918 

1951 

997493 

23 

032425 

1974 

e'67575 

51 

10 
11 

031089 

1947 

997480 

23 
23 

033609 
9.034791 

1969 

966391 

50 
49 

9.032257 

1941 

9.997466 

1964 

10.965209 

12 

033421 

1936 

997452 

23 

035969 

1958 

964031 

48 

13 

034582 

1930 

997439 

23 

0.37144 

1953 

962856 

47 

14 

035741 

1925 

997425 

23 

038316 

1948 

961684 

46 

15 

036896 

1920 

997411 

23 

039485 

1943 

960515 

45 

16 

038048 

1915 

997397 

23 

040651 

1938 

959349 

44 

17 

039197 

1910 

997383 

23 

041813 

1933 

9.58187 

43 

18 

040342 

1905 

997369 

23 

042973 

1928 

957027 

42 

19 

041485 

1899 

997355 

23 

044130 

1923 

955870 

41 

20 
21 

042625 

1894 

997341 
9.997327 

23 

24 

045284 
9.046434 

1918 

9.54716 

40 
39 

9.043762 

1889 

1913 

10.9,53566 

22 

044895 

1884 

997313 

24 

047582 

1908 

952418 

38 

23 

046026 

1879 

997299 

24 

048727 

1903 

951273 

37 

24 

047154 

1875 

997285 

24 

049869 

1898 

950131 

36 

25 

048279 

1870 

997271 

24 

051008 

1893 

948992 

35 

26 

049400 

1865 

997257 

24 

052144 

1889 

947856 

34 

27 

050519 

1860 

997242 

24 

053277 

1884 

946723 

33 

28 

051635 

1855 

997228 

24 

054407 

1879 

945593 

32 

29 

052749 

1850 

997214 

24 

055535 

1874 

944465 

31 

30 

053859 

1845 

997199 

24 

056659 

1870 

943341 

30 

31 

054966 

1841 

9.997185 

24 

9.057781 

1865 

10.942219 

29 

32 

056071 

1836 

997170 

24 

058900 

1869 

941100 

28 

33 

0.57172 

1831 

997156 

24 

060016 

1855 

939984 

27 

34 

058271 

1827 

997141 

24 

061130 

1851 

938870 

26 

35 

059307 

1822 

997127 

24 

062240 

1846 

937760 

25 

36 

060460 

1817 

997112 

24 

063348 

1842 

936652 

24 

37 

061551 

1813 

997098 

24 

064453 

1837 

935547 

23 

38 

002639 

1808 

997083 

25 

065556 

1833 

934444 

22 

39 

063724 

1804 

997068 

25 

066655 

1828 

933345 

21 

40 
41 

064806 
9.065885 

1799 

997053 

25 

25 

067752 

1824 

932248 

20 
19 

1794 

9.997039 

9.068846 

1819 

10.931154 

42 

066962 

1790 

997024 

25 

069938 

1815 

930062 

18 

43 

068036 

1786 

997009 

25 

071027 

1810 

928973 

17 

44 

069107 

1781 

996994 

25 

072113 

1806 

927887 

16 

45 

070176 

1777 

996979 

25 

073197 

1802 

926803 

15 

46 

071242 

1772 

996964 

25 

074278 

1797 

925722 

14 

47 

072306 

1768 

996949 

25 

075356 

1793 

924644 

13 

48 

073366 

1763 

996934 

25 

076432 

1789 

923568 

12 

49 

074424 

1759 

996919 

25 

077505 

1784 

922495 

11 

50 
51 

075480 
9.076.533 

1755 

996904 

25 
25 

078576 

1780 
1776 

921424 

10 
9 

1  1750 

9.996889 

9.079644 

10.920356 

52 

077583 

1746 

996874 

25 

1   080710 

1772 

919290 

8 

53 

078031 

1742 

996858 

25 

081773 

1767 

9  J  8227 

7 

54 

079676 

1738 

996843 

25 

082833 

1763 

917167 

6 

55 

080719 

1733 

996828 

25 

083891 

1759 

916109 

5 

56 

081759 

i  1729 

996812 

26 

084947 

1755 

915053 

4 

57 

08279V 

1  1725 

996797 

26 

086000 

1751 

914000 

3 

58 

083832 

1721 

996782 

26 

087050 

1747 

912950 

2 

59 

084864 

1717 

996766 

26 

088098 

1  1743 

911902 

1 

60 

085894 

1713 

996751 

26 

089144 

!  1738 

9108.56 

0 

Cosine 

Sine   1 

1   «;o.au{^. 

1 

1   Tang.   1 M.  | 

83  Degrees. 


siivEs  AND  TANGLKTs.  ^7  Dcgrees.) 

25 

M. 

Sine 

D. 

Cosine   1  D. 

Tang.   1 

D. 

Cotang.   (    1 

0 

9.085894 

1713 

9.996751 

26 

9.089144 

1738 

10.910850 

-60 

1 

086922 

1709 

996735 

26 

090187 

1734 

909813 

59 

2 

087947 

1704 

996720 

26 

091228 

1730 

908772 

58 

3 

088970 

1700 

996704 

26 

092266 

1727 

907734 

57 

4 

089990 

1696 

996688 

26 

093302 

1722 

906698 

56 

6 

091008 

1692 

996673 

26 

094336 

1719 

905664 

55 

6 

092024 

1688 

996657 

26 

095367 

1715 

9046'}3 

54 

7 

093037 

1684 

996641 

26 

096395 

1711 

903605 

53 

8 

094047 

1680 

996625 

26 

097422 

1707 

902578 

52 

9 

095056 

1676 

996610 

26 

098446 

1703 

901554 

51 

10 
11 

096062 

1673 

996594 
9.996578 

26 

27 

099468 
9.100487 

1699 

900532 

50 
49 

9.097065 

1668 

1695 

10.899513 

12 

098066 

1665 

996562 

27 

101504 

1691 

898496 

48 

13 

099065 

1661 

996546 

27 

102519 

1687 

897481 

47 

14 

J00062 

1657 

996530 

27 

103532 

1684 

896468 

46 

15 

101056 

1653 

996514 

27 

104542 

1680 

895458 

45 

16 

102048 

1649 

996498 

27 

105550 

1676 

894450 

44 

17 

103037 

1645 

996482 

27 

106556 

1672 

893444 

43 

18 

104025 

1641 

996465 

27 

107559 

1669 

892441 

42 

19 

105010 

1638 

996449 

27 

108560 

1665 

891440 

41 

20 

21 

105992 

1634 
1630 

996433 

27 

27 

109559 

1661 

890441 

40 
39 

9.106973 

9.996417 

9.110556 

1658 

10.889444 

22 

107951 

1627 

990400 

27 

111551 

1654 

888449 

38 

23' 

108927 

1023 

996384 

27 

112543 

1650 

887457 

37 

24 

109901 

1619 

996368 

27 

113533 

1646 

886467 

36 

25 

110873 

1616 

996351 

27 

114521 

1643 

885479 

35 

26 

111842 

1612 

996335 

27 

115507 

1639 

884493 

34 

27 

112809 

1608 

996318 

27 

116491 

1636 

883509 

33 

28 

113774 

1605 

996302 

28 

117472 

1632 

882528 

32 

29 

114737 

1601 

996285 

28 

118452 

1629 

881548 

31 

30 
31 

115698 

1597 
1594 

996269 
9.996252 

28 
28 

119429 

1625 

880571 
10.879.596 

30 
29 

9.116656 

9.120404 

1622 

32 

117613 

1590 

996235 

28 

121377 

1618 

878623 

28 

33 

118567 

1587 

996219 

28 

122348 

1615 

877652 

27 

34 

119519 

1583 

996202 

28 

123317 

1611 

876683 

26 

35 

120469 

1580 

996185 

28 

124284 

1607 

875716 

25 

36 

121417 

1576 

996168 

28 

125249 

1604 

874751 

24 

37 

122362 

1573 

996151 

28 

126211 

1601 

873789 

23 

38 

123306 

1569 

996134 

28 

127172 

1597 

872828 

22 

39 

124248 

1566 

996117 

28 

128130 

1594 

871870 

21 

40 
41 

125187 

1562 

996100 
9.996083 

28 
29 

129087 
9.130041 

1591 

870913 

20 
19 

9.126125 

1559 

1587 

10.8699.59 

42 

127060 

1556 

996066 

29 

130994 

1584 

869006 

18 

43 

127993 

1552 

996049 

29 

131944 

1581 

868056 

17 

44 

128925 

1549 

996032 

29 

132893 

1577 

867107 

16 

45 

129854 

1545 

996015 

29 

1338.39 

1574 

866161 

15 

46 

130781 

1542 

995998 

29 

1.34784 

1571 

865216 

14 

47 

131706 

1539 

995980 

29 

135726 

1567 

864274 

13 

48 

.  132630 

1535 

995963 

29 

136667 

1564 

863333 

12 

49 

133551 

1532 

995946 

29 

137605 

1561 

862395 

11 

50 

51 

134470 
9.135387 

1529 
1525 

995928 
9.995911 

29 
29 

138542 

1558 

861458 

10 
9 

9.139476 

1555 

10.860524 

52 

136303 

1522 

99.5894 

29 

140409 

1561 

859591 

8 

53 

137216 

1519 

995876 

29 

141340 

1548 

858660 

7 

54 

138128 

1516 

995859 

29 

142269 

1545 

857731 

6 

55 

139037 

1512 

995841 

29 

143196 

1542 

856804 

6 

56 

139944 

1509 

995823 

29 

144121 

1539 

855879 

4 

57 

140850 

1506 

995806 

29 

145044 

1535 

854956 

3 

58 

141754 

1503 

995788 

29 

145966 

1532 

854034 

2 

59 

142655 

1500 

995771 

29 

146885 

1.529 
1526 

853115 

1 

60 

143555 

1496 

995753 

29 

147803 

852197 

0 

3 

Cosine  { 

1 

bine   | 

roiai.t.  1 

.! 

Tung    |M;| 

Cc 


Degreei. 


26 

(8 

Degrees.;  a  table  op  logarithmic 

Ti 

Sine 

D.   1 

Cosine   |  D.  | 

Tan^j.   1 

D.   1 

Ct)tajig.  1   1 

0 

9.143556 

1496 

9.996753 

30- 

9.147803 

1526 

10.8621971  60  1 

1 

144453 

1493 

995735 

30 

148718 

1523 

851282 

59 

2 

145349 

1490 

995717 

30 

149632 

1520 

850368 

68 

3 

146243 

1487 

995699 

30 

150544 

1517 

849466 

67 

4 

147136 

1484 

995681 

30 

161454 

1514 

848546 

66 

5 

148026 

1481 

995664 

30 

152363 

1511 

847637 

55 

G 

148915 

1478 

995646 

30 

153^9 

1508 

846731 

54 

7 

149802 

1475 

995628 

30 

164174 

1505 

845826 

53 

8 

150686 

1472 

995610 

3a 

155077 

1502 

844923 

52 

9 

151569 

1469 

995591 

30 

155978 

1499 

844022 

21 

10 
11 

152451 

1466 

995573 

30 
30 

156877 
9.157775 

1496 

843123 

50 
49 

9  153330 

1463 

9.995655 

1493 

10.842225 

12 

154208 

1460 

995537 

30 

158671 

1490 

841329 

48 

13 

155083 

1467 

995519 

30 

159565 

1487 

840435 

47 

14 

155957 

1454 

995501 

31 

160467 

1484 

839543 

46 

15 

156830 

1451 

995482 

31 

161347 

1481 

838653 

46 

16 

167700 

1448 

996464 

31 

162236 

1479 

837764 

44 

17 

158569 

1445 

995446 

31 

163123 

1476 

836877 

43 

18 

169435 

1442 

995427 

31 

164008 

1473 

835992 

42 

19 

100301 

1439 

996409 

31 

164892 

1470 

835108 

41 

20 
21 

161164 
9.162026 

1436 
1433 

996390 

31 
31 

165774 

1467 

834226 

40 
39 

9.996372 

9.166654 

1464 

10.833346 

22 

162885 

1430 

995353 

31 

167532 

1461 

832468 

38 

23 

163743 

1427 

99.5334 

31 

168409 

1468 

831591 

37 

24 

164600 

1424 

995316 

31 

169284 

1466 

830716 

36 

25 

165454 

1422 

995297 

31 

170157 

1453 

829843 

35 

26 

166307 

1419 

995278 

31 

171029 

1460 

828971 

34 

27 

167159 

1416 

995260 

31 

171899 

1447 

828101 

33 

28 

168008 

1413 

995241 

32 

172767 

1444 

827233 

32 

29 

168856 

1410 

995222 

32 

173634 

1443 

826366 

31 

30 
31 

169702 
9.170547 

1407 

995203 

32 
32 

174499 

1439 

825501 

30 

29 

1405 

9.996184 

9,175362 

1436 

10.824638 

32 

171389 

1402 

995165 

32 

176224 

1433 

823776 

28 

33 

172230 

1399 

995146 

32 

177084 

1431 

822916 

27 

34 

173070 

1396 

995127 

32 

177942 

1428 

822058 

26 

35 

173908 

1394 

996108 

32 

178799 

1426 

821201 

25 

36 

174744 

1391 

996089 

32 

179665 

1423 

820346 

24 

37 

175578 

1388 

995070 

32 

180508 

1420 

819492 

23 

38 

176411 

1386 

996051 

32 

-181360 

1417 

818640 

22 

39 

177242 

1383 

995032 

32 

182211 

1416 

817789 

21 

40 
41 

178072 

1380 
1377 

995013 

32 
32 

183059 

1412 

816941 

20 
19 

9.178900 

9.994993 

9.183907 

1409 

10.816093 

42 

179726 

1374 

994974 

32 

184752 

1407 

815248 

18 

43 

180551 

1372 

994955 

32 

185597 

1404 

814403 

17 

44 

181374 

1369 

994935 

32 

186439 

1402 

813561 

16 

45 

182196 

1366 

994916 

33 

187280 

1399 

812720 

15 

46 

183016 

1364 

994896 

33 

188120 

1396 

811880 

14 

47 

183834 

1361 

994877 

33 

188958 

1393 

811042 

13 

48 

184651 

1369 

994857 

33 

189794 

1391 

810206 

12 

49 

185466 

1356 

994838 

33 

190629 

1389 

809371 

11 

50 
51 

186280 

1353 

994818 

33 
33 

191462 
9.192294 

1386 
1384 

808538 
10.807706 

10 
9 

9.187092 

1351 

9.994798 

52 

187903 

1348 

994779 

33 

193124 

1381 

80C876 

8 

53 

188712 

1346 

994759 

33 

193953 

1379 

80604? 

7 

54 

189519 

1343 

994739 

33 

194780 

1376 

805220 

6 

55 

190325 

1341 

994719 

33 

195606 

1374 

804394 

6 

56 

191130 

1338 

994700 

33 

196430 

1371 

803570 

4 

57 

191933 

1336 

994680 

33 

197253 

1369 

802747 

3 

58 

192734 

1333 

994660 

33 

198074 

1366 

801926 

2 

59 

193534 

1330 

994640  33 

198894 

1364 

801106 

7 

60 

194332 

1328 

994620  33 

199713 

1361 

800287 

0 

1  Cosine 

1 

1   «'"«   1 

1  Cotang. 

1   Tang.  \  M. 

81  Degreea. 


SINES  AND  TANGENTS.  (9  Degrees.) 

27 

M. 

Sine   j 

D.   1 

Cosine   |  D. 

Tang. 

D. 

Cotang.   1 

0 

9.194332 

1328 

9.994620 

33 

9,199713 

1361 

10.800287  60 

1 

1951S9 

1326 

994600 

33 

200629 

1359 

799471 

59 

2 

195926 

1323 

994580 

33 

201345 

1356 

798655 

5S 

3 

196719 

1321 

994560 

34 

202159 

1354 

797841 

57 

4 

197511 

1318 

994540 

34 

202971 

1352 

797029 

56 

5 

198302 

1316 

994519 

34 

203782 

1349 

796218 

55 

6 

199091 

1313 

994499 

34 

204592 

1347 

796408 

64 

7 

199879 

1311 

994479 

34 

205400 

1345 

794600 

53 

8 

200666 

1308 

994459 

34 

206207 

1342 

793793 

52 

9 

201461 

1306 

994438 

34 

2^7013 

1340 

792987 

51 

10 
11 

202234 

1304 
1301 

994418 

34 
34 

207817 

1338 

792183 

50 

49 

9,203017 

9.994397 

9,208619 

1335 

10,791381 

12 

203797 

1299 

994377 

34 

209420 

1333 

790580 

48 

13 

204577 

1296 

994357 

34 

210220 

1331 

789780 

47 

14 

205354 

1294 

994336 

34 

211018 

1328 

788982 

46 

15 

206131 

1292 

994316 

34 

211815 

1326 

788185 

45 

16 

206906 

1289 

994295 

34 

212611 

1324 

787389 

44 

17 

207679 

1287 

994274 

36 

213405 

1321 

786595 

43 

18 

208452 

1285 

994254 

35 

214198 

1319 

785802 

42 

19 

209222 

1282 

994233 

35 

214989 

1317 

785011 

41 

20 
21 

209992 
9,210760 

1280 

994212 

35 
35 

215780 
9,216668 

1315 

784220 

40 
39 

1278 

9,994191 

1312 

10,783432 

22 

211526 

1275 

994171 

35 

217366 

1310 

782644 

38 

23 

212291 

1273 

994160 

36 

218142 

1308 

781858 

37 

24 

213055 

1271 

994129 

35 

218926 

1305 

781074 

36 

25 

213818 

1268 

994108 

35 

219710 

1303 

780290 

35 

26 

214579 

1266 

994087 

35 

220492 

1301 

779608 

34 

27 

215338 

1264 

994066 

35 

221272 

1299 

778728 

33 

28 

216097 

1261 

994045 

35 

222052 

1297  , 

777948 

32 

^ 

^16854 

1259 

994024 

35 

•222830 

1294 

777170 

31 

30 
31 

217609 

1257 
1255 

994003 

35 
35 

223606 

129,2 

776394 

30 
29 

9-218363 

9.993981 

9.224382 

1290 

10.775618 

32 

219116 

1253 

993960 

35 

225156 

1288 

774844 

28 

33 

-215868 

1250 

993939 

35 

226929 

1286 

774071 

27 

34 

22D618 

1248 

993918 

35 

226700 

1284 

773300 

26 

35 

221367 

1246 

993896 

36 

227471 

1281 

772529 

25 

36 

222115 

1244 

993875 

36 

228239 

1279 

771761 

24 

37 

222861 

1242 

993854 

36 

229007 

1277 

770993 

23 

38 

223606 

1239 

993S32 

36 

229773 

1275 

770227 

22 

39 

224349 

1237 

993811 

36 

230539 

1273 

769461 

21 

40 
41 

225092 

1235 

993789 

36 
36 

231302 

1271 

768698 

20 
19 

9.225833 

1233 

9.993768 

9-232065 

1269 

10.767935 

42 

226573 

1231 

993746 

36 

232826 

1267 

767174 

18 

43 

227311 

1228 

993725 

36 

233586 

1265 

766414 

17 

44 

228048 

1220 

993703 

36 

234345 

1262 

766665 

16 

45 

228784 

1224 

993681 

36 

235103 

1260 

764897 

15 

46 

229518 

1222 

993660 

36 

235859 

1258 

764141 

14 

47 

230252 

1220 

993638 

36 

236614 

1256 

763386 

13 

48 

230984 

1218 

993616 

36 

237368 

1254 

762632 

12 

49 

231714 

1216 

993594 

37 

238120 

1252 

761880 

11 

50 
51 

232444 

1214 

993572 

37 
37 

238872 

1250 

76112S 
10.760378 

10 
9 

9.233172 

1212 

9.993550 

.9.239622 

1248 

52 

233899 

1209 

993528 

37 

240371 

1246 

769629 

8 

63 

234625 

1207 

993506 

37 

241118 

1244 

758882 

7 

54 

235349 

1205 

993484 

37 

241865 

1242 

758135 

6 

55 

236073 

1203 

993462 

37 

242610 

1240 

757390 

5 

56 

236795 

1201 

993440 

37 

243354 

1238 

756646 

4 

57 

237515 

1199 

993418 

37 

244097 

1236 

766903 

3 

58 

238235 

1197 

993396 

37 

244839 

1234 

7.55161 

2 

59 

238958 

1195 

993374 

37 

246678 

1232 

754421 

1 

60 

230670 

1193 

993351 

37 

246319 

1230 

753681 

0 

Ll 

Cosine 

i 

Sine   I 

Cotang. 

Tang.   M.  | 

30  Degrees. 


58 

(10  Degr 

ees. )    A 

TABLE  OP  LOGARITHMIC^ 

M. 

Sine 

1   D. 

Cosine   1  D. 

Tang. 

1   D. 

1   Cotang.  1   1 

0 

9.239670 

1193 

9.993351 

37 

9.246319 

1230 

10.753681 

60 

1 

240386 

1191 

993329 

37 

247057 

1228 

752943 

59 

2 

241101 

1189 

993307 

37 

247794 

1226 

752206 

68 

3 

241814 

1187 

993285 

37 

248530 

1224 

751470 

57 

4 

242526 

1185 

9S3262 

37 

249264 

1222 

750736 

56 

6 

243237 

1183 

993240 

37 

249998 

1220 

760002 

55 

6 

243947 

1181 

993217 

38 

2507.30 

1218 

749270 

54 

7 

244656 

1179 

993195 

38 

261461 

1217 

748639 

53 

8 

245363 

1177 

993172 

38 

252191 

1215 

747809 

52 

9 

246069 

1175 

993149 

38 

252920 

1213 

747080 

51 

10 
11 

246775 

1173 
1171 

993127 

38 

38 

263648 

1211 

746362 

50 
49 

9.247478 

9.993104 

9.254374 

1209 

10.746626 

12 

248181 

1169 

993081 

38 

256100 

1207 

744900 

48 

13 

248883 

1167 

993059 

38 

266824 

1205 

744176 

47 

14 

249583 

1165 

993036 

38 

266547 

1203 

743463 

46 

15 

250282 

1163 

993013 

38 

257269 

1201 

742731 

45 

16 

250980 

1161 

992990 

38 

2.57990 

1200 

742010 

44 

17 

251677 

1159 

992967 

38 

258710 

1198 

741290 

43 

18 

252373 

1158 

992944 

38 

259429 

1196 

740571 

42 

19 

253067 

1156 

992921 

38 

260146 

1194 

739854 

41 

20 

21 

253761 
9.254453 

1154 
1152 

992898 

38 

38 

260863 
9.261578 

1192 

739137 

40 
39 

9.992875 

1190 

10.738422 

22 

255144 

1160 

992852 

38 

262292 

1189 

737708 

38 

23 

255834 

1148 

992829 

39 

263005 

1187 

736995 

37 

24 

266523 

1146 

992806 

39 

263717 

1185 

736283 

36 

25 

257211 

1144 

992783 

39 

264428 

1183 

735572 

35 

26 

257898 

1142 

992759 

39 

265138 

1181 

734862 

34 

27 

258583 

1141 

993736 

39 

266847 

1179 

734153 

33 

28 

259268 

1139 

992713 

39 

266565 

1178 

733445 

32 

29 

259951 

1137 

992690 

39 

267261 

1176 

732739 

31 

30 
31 

260633 

1135 
1133 

992666 

39 
39 

267967 

1174 

732033 

30 
29 

9.261314 

9.992643 

9.268671 

1172 

10.731329 

32 

261994 

1131 

992619 

39 

269376 

1170 

730625 

28 

33 

262673 

1130 

992596 

39 

270077 

1169 

729923 

27 

34 

263351 

1128 

992572 

39 

270779 

1167 

729221 

26 

35 

264027 

1126 

992549 

39 

271479 

1165 

728521 

25 

36 

264703 

1124 

992526 

39 

272178 

1164 

727822 

24 

37 

265377 

1122 

992501 

39 

272876 

1162 

727124 

23 

38 

266051 

1120 

992478 

40 

273573 

1160 

726427 

•22 

39 

266723 

1119 

992464 

40 

274269 

1158 

726731 

21 

40 
41 

267395 
9.268065 

1117 

992430 

40 
40 

274964 

1167 

726036 

20 
19 

1115 

9.992406 

9.275658 

1155 

10.724342 

42 

268734 

1113 

992382 

40 

276351 

1153 

723649 

18 

43 

269402 

1111 

992359 

40 

277043 

1151 

722967 

17 

44 

270069 

1110 

992335 

40 

277734 

1150 

722266 

16 

45 

270735 

1108 

992311 

40 

278424 

1148 

721676 

15 

46 

271400 

1106 

992287 

40 

279113 

1147 

720887 

14 

47 

272064 

1105 

992263 

40 

279801 

1145 

720199 

13 

48 

272726 

1103 

992239 

40 

280488 

1143 

719512 

12 

49 

273388 

1101 

992214 

40 

281174 

1141 

718826 

11 

50 
51 

274049 
9.274708 

1099 

992190 

40 
40 

281858 
9.282642 

1140 

718142 

10 
9 

1098 

9.992106 

1138 

10.717458 

52 

275367 

1096 

992142 

40 

283225 

1136 

716775 

8 

53 

276024 

1094 

992117 

41 

283907 

1135 

716093 

7 

64 

276681 

1092 

992093 

41 

284588 

1133 

715412 

6 

55 

277337 

1091 

992069 

41 

285268 

1131 

714732 

5 

56 

277991 

1089 

992044 

41 

286947 

1130 

714053 

4 

57 

278644 

1087 

992020 

41 

286624 

1128 

713376 

3 

58 

279297 

1086 

991996 

41 

287301 

1126 

712699 

2 

59 

279948 

1084 

991971 

41 

287977 

1125 

712023 

1 

60 

280599 

1082 

991947 

41 

288662 

1123 

711348 

0 

1 

Cosine  | 

Sine  .   1    1 

Colang.  1 

1 

Tang.   1  M.  j 

79  Degrees, 


SINES  AND  TANGENTS 

(11  Degrees. 

) 

29 

M. 

1   Sine 

1   D. 

1  Co^ne      1  D. 

1   Tang.   1   D. 

1   Cotang.   1   1 

T 

9.280599 

1082 

9.991947 

41 

9.288652 

1123 

10.7113481  60 1 

1 

281248 

1081 

991922 

41 

289326 

1123 

710674 

59 

2 

281897 

1079 

991897 

41 

289999 

1120 

710001 

58 

3 

282544 

1077 

991873 

41 

290671 

1118 

709329 

57 

4 

283190 

1076 

991848 

41 

291342 

1117 

708658 

56 

5 

283836 

1074 

991823 

41 

292013 

1115 

707987 

55 

& 

284480 

1072 

991799 

41 

292682 

1114 

707318 

54 

7 

285124 

1071 

991774 

42 

293350 

1112 

706650 

53 

8 

285766 

1069 

991749 

42 

294017 

1111 

705983 

52 

9 

286408 

1067 

991724 

42 

294684 

1109 

705316 

51 

10 
11 

287048 
9,287687 

1066 

991699 
9.991674 

42 

42 

295349 

1107 

704651 

50 
49 

1064 

9.296013 

1106 

10.703987 

12 

288326 

1063 

991649 

42 

296677 

1104 

703323 

48 

13 

288964 

1061 

991624 

42 

297339 

1103 

702661 

47 

14 

289600 

1059 

991599 

42 

298001 

1101 

701999 

46 

15 

290236 

1058 

991574 

42 

298662 

1100 

701338 

45 

16 

290870 

1056 

991549 

42 

299322 

1098 

700678 

44 

17 

291504 

1054 

991524 

42 

299980 

1096 

700020 

43 

18 

293137 

1053' 

991498 

42 

300638 

1095 

699362 

42 

19 

292768 

1051 

991473 

42 

301295 

1093 

698705 

41 

20 

293399 

1050 

991448 

42 

301951 

1092 

698049 

40 

21 

9.294029 

1048 

9.991422 

42 

9.302607 

1090 

10.697393 

39 

22 

294658 

1046 

991397 

42 

303261 

1089 

696739 

38 

23 

295280 

1045 

991372 

43 

303914 

1087 

696086 

37 

24 

295913 

1043 

991346 

43 

304567 

1086 

695433 

36 

25 

296539 

1042 

991321 

43 

305218 

1084 

694782 

35 

26 

297164 

1040 

991295 

43 

305869 

1083 

694131 

34 

27 

297788 

1039 

991270 

43 

306519 

1081 

693481 

33 

28 

298412 

1037 

991244 

43 

307168 

1080 

692832 

32 

29 

299034 

1036 

991218 

43 

307815 

1078 

692185 

31 

30 
31 

299655 
9.300276 

1034 

991193 
9.991167 

43 

43 

308463 
9.309109 

1077 

691537 

30 

29 

1032 

1075 

10.690891 

32 

300895 

1031 

991141 

43 

309754 

1074 

690246 

28 

33 

301514 

1029 

991115 

43 

310398 

1073 

689602 

27 

34 

302132 

1028 

991090 

43 

811042 

1071 

688958 

26 

35 

302748 

1026 

991064 

43 

311685 

1070 

688315 

25 

36 

303364 

1025 

991038 

43 

312327 

1068 

687673 

24 

37 

303979 

1023 

991012 

43 

312967 

1067 

687033 

23 

38 

304593 

1022 

990986 

43 

313608 

1005 

686392 

22 

39 

305207 

1020 

990960 

43 

314247 

1064 

685753 

21 

40 

305819 

1019 

990934 

44 

314885 

1062 

685115 

20 

41 

9.806430 

1017 

9.990908 

44 

9.315523 

1061 

10.684477 

19 

42 

307041 

1016 

990882 

44 

316159 

1060 

683841 

18 

43 

307650 

1014 

990855 

44 

316795 

1058 

683205 

17 

44 

308259 

1013 

990829 

44 

317430 

1057 

683570 

16 

45 

308867 

1011 

990803 

44 

318064 

1055 

681936 

li 

46 

309474 

1010 

990777 

44 

318697 

1054 

681303 

47 

310080 

1008 

990750 

44 

319329 

1053 

680671 

13 

48 

310685 

1007 

990724 

44 

319961 

1051 

680039 

12 

49 

311289 

1005 

990697 

44 

320592 

1050 

679408 

11 

50 

311893 

1004 

990671 

44 

321222 

1048 

678778 

10 

51 

9.312495 

1003 

9.990644144 

9.321851]  1047 

10.678149 

9 

62 

313097 

1001 

990618144 

322479   1045 

677521 

8 

53 

313698 

1000 

990591  44 

323106  1044 

676894 

7 

54 

314297 

•998 

9905651  44 

323733   1043 

676267 

6 

55 

314897 

997 

990538:  44 

324358!  1041 

675642 

5 

56 

315495 

996 

9905  ir  45 

324983  1040 

675017 

4 

57 

316092 

994 

990485  45 

325607;  1039 

6743931  3 

58 

316689 

993 

990458  45 

326231'  1037 

6737691  2 

59 

317284 

991 

990431  45 

326853  1036 

673147!  1 

60 

317879 

990 

990404  45 

327475  1035 

672525*  0 



Cosine  | 

Bine    | 

Uotaiig.  j 

Tang,   j 

Cc 


78  Degrees. 


30 

(12  Degrees.)  a 

TABLE  OF  LOGAEITHMIC 

M. 

1    Sine 

1   D. 

1   Cosine   1  D. 

1   Tang. 

1   D. 

1   Cota.ig.  1   1 

T 

9.317879 

990 

9.990404 

45 

9.327474 

1035 

10.672526 

60 

1 

318473 

988 

990378 

45 

328095 

1033 

671905 

59 

2 

319066 

987 

990351 

45 

328715 

1032 

671285 

58 

3 

319658 

986 

990324 

45 

329334 

1030 

670666 

57 

4 

320249 

984 

990297 

45 

329953 

1029 

670047 

56 

5 

320840 

983 

990270 

45 

330570 

1028 

669430 

55 

6 

,321430 

982 

990243 

45 

331187 

1026 

668813 

54 

7 

322019 

980 

990215 

45 

331803 

1025 

668197 

53 

8 

322607 

979 

990188 

45 

332418 

1024 

667582 

52 

9 

323194 

977 

990161 

45 

333033 

1023 

666967 

51 

10 
11 

323780 

976 

990134 

45 
46 

333646 

1021 

666354 

50 
49 

9.324366 

975 

9.990107 

9.334259 

1020 

10.665741 

12 

324950 

973 

990079 

46 

334871 

1019 

665129 

48 

13 

325534 

972 

990052 

46 

335482 

1017 

664518 

47 

14 

326117 

970 

990025 

46 

336093 

1016 

663907 

46 

15 

326700 

969 

98-^997 

46 

336702 

1015 

663298 

45- 

16 

327281 

968 

98997a 

46 

337311 

1013 

662689 

44 

17 

327862 

966 

989942 

46 

337919 

1012 

662081 

43 

18 

328442 

965 

989915 

46 

338527 

1011 

661473 

42 

19 

329021 

964 

989887 

46 

339133 

1010 

660867 

41 

20 
21 

329599 

962 
961 

989860 

46 
46 

339739 
9.340344 

1008 
1007 

660261 

40 
39 

9.330176 

9.989832 

10.659656 

22 

330753 

960 

989804 

46 

340948 

1006 

659052 

38 

23 

331329 

958 

989777 

46 

341552 

1004 

658448 

37 

24 

331903 

957 

989749 

47 

342155 

1003 

657845 

36 

25 

332478 

956 

989721 

47 

342757 

1002 

657243 

35 

26 

333051 

954 

989693 

47 

343358 

1000 

656642 

34 

27 

333624 

953 

989665 

47 

843958 

999 

656042 

33 

28 

334195 

952 

989637 

47 

344558 

998 

655442 

32 

29 

334766 

950 

989609 

47 

345157 

997 

654843 

31 

30 
31 

335337 

949 

989582 
9.989553 

47 

47 

345755 
9.346353 

996 

654245 

30 
29 

9.335906 

948 

994 

10.653647 

32 

336475 

946 

989525 

47 

346949 

993 

653051 

28 

33 

337043 

945 

989497 

47 

347545 

992 

652455 

27 

34 

337610 

944 

989469 

47 

348141 

991 

651859 

26 

35 

338176 

943 

989441 

47 

348735 

990 

651265 
650671 

25 

36 

338742 

941 

989413 

47 

349329 

988 

24 

37 

339306 

940 

989384 

47 

349922 

987 

650078 

23 

38 

339871 

939 

989356 

47 

350514 

986 

649480 

22 

39 

340434 

937 

989328 

47 

351106 

985 

648894 

21 

40 
41 

340996 

936 

989300 
9.989271 

47 
47 

351697 

983 

982 

648303 

20 
19 

9.341558 

935 

9.352287 

10.647713 

42 

342119 

934 

989243 

47 

352876 

981 

647124 

18 

43 

342679 

932 

989214 

47 

353465 

980 

646535 

17 

44 

343239 

931 

989186 

47 

354053 

979 

645947 

16 

45 

343797 

930 

989157 

47 

354640 

977 

645360 

15 

46 

344355 

929 

989128 

48 

355227 

976 

644773 

14 

47 

344912 

927 

989100 

48 

355813 

975 

644187 

13 

48 

345469 

926 

989071 

48 

356398 

974 

643602 

12 

49 

346024 

925 

989042 

48 

356982 

973 

643018 

11 

50 
51 

346579 

924 

989014 
9.988985 

48 
48 

357566 
9.358149 

971 
970 

642434 

10 
9 

9.347134 

922 

10.641851 

52 

347687 

921 

988956 

48 

358731 

969 

641269 

8 

53 

348240 

920 

988927 

48 

359313 

968 

640687 

7 

54 

348792 

919 

988898 

48 

359893 

967 

640107 

6 

55 

349343 

917 

988869 

48 

360474 

966 

639526 

5 

56 

349893 

916 

988840 

48 

361053 

965 

638947 

4 

57 

350443 

915 

988811 

49 

361632 

963 

638368 

3 

58 

350992 

914 

988782 

49 

362210 

962 

637790 

2 

59 

351540 

913 

988753 

49 

362787 

961 

637213 

1 

60 

352088 

911 

988724 

49 

363364 

960 

636636 

0 

l: 

Cosine 

1   Sine   1 

Colaiig. 

Tang   1 M.  1 

77  Degrees. 


SINES  AND  TANGENTS.  (13  Degrees.) 

31 

J4. 

1   Sine 

D. 

1  Cosine   1  I). 

1   Tanp. 

1   D. 

1   Ootang.   1   1 

^ 

9.3520S8 

911 

9.988724 

49 

9.363364 

960 

10.636636 

60 

1 

352635 

910 

988695 

49 

363940 

959 

636060 

59 

2 

353181 

909 

988666 

49 

364515 

958 

635485 

58 

3 

353726 

908 

988636 

49 

365090 

957 

634910 

57 

4 

354271 

907 

988607 

49 

365664 

955 

6343361  56  I 

5 

354815 

905 

988578 

49 

366237 

954 

633763 

55 

6 

355358 

904 

988548 

49 

366810 

953 

633190 

54 

7 

355901 

903 

988519 

49 

367382 

952 

632618 

53 

8 

356443 

902 

988489 

49 

367953 

951 

632047 

52 

9 

356984 

901 

988460 

49 

368524 

950 

631476 

51 

10 
11 

357524 

899 

988430 
9.988401 

49 
49 

369094 

949 

630906 

50 
49 

9.358064 

898 

9.369663 

948 

10.630337 

12 

358603 

897 

988371 

49 

370232 

946 

629768 

48 

13 

359141 

896 

988342 

49 

370799 

945 

629201 

47 

14 

359678 

895 

988312 

50 

371367 

944 

628633 

46 

15 

360215 

893 

988282 

50 

371933 

943 

623067 

45 

16 

360752 

892 

988252 

50 

372499 

942 

627501 

44 

17 

361287 

891 

988223 

50 

373064 

941 

626936 

43 

18 

361822 

890 

988193 

/50 

373629 

940 

626371 

42 

19 

362356 

889 

988163 

50 

374193 

939 

625807 

41 

20 
21 

362889 
9.363422 

888 

988133 

50 
50 

374756 

938 

625244 

40 
39 

887 

9.988103 

9.375319 

937 

10.624681 

22 

363954 

885 

988073 

50 

375881 

935 

624119 

38 

23 

364485 

884 

988043 

50 

376442 

934 

623558 

37 

24 

365016 

883 

988013 

50 

377003 

933 

622997 

36 

25 

365546 

882 

987983 

50 

377563 

932 

622437 

35 

26 

366875 

881 

987953 

50 

378122 

931 

621878 

34 

27 

366604 

880 

987922 

50 

378681 

930 

621319 

33 

28 

367131 

879 

987892 

50 

379239 

929 

620761 

32 

29 

367659 

877 

987862 

50 

379797 

928 

620203 

31 

30 
31 

368185 

876 

987832 
9.987801 

51 
51 

380354 

927 

619046 

30 

29 

9.368711 

875 

9.380910 

926 

10.619090 

32 

369236 

874 

987771 

51 

381466 

925 

618534 

28 

33 

369761 

873 

987740 

51 

382020 

924 

617980 

27 

34 

370285 

872 

987710 

51 

382575 

923 

617425 

26 

35 

370808 

871 

987679 

51 

383129 

922 

616871 

25 

36 

371330 

870 

987649 

51 

383682 

921 

616318 

24 

37 

.371852 

869 

987618 

51 

384234 

920 

615766 

23 

38 

372373 

867 

987588 

51 

384786 

919 

615214 

22 

39 

372894 

866 

987557 

51 

385337 

918 

614663 

21 

40 
41 

373414 

865 

987526 
9.987496 

51 

51 

385888 

917 

614112 

20 
19 

9.373933 

864 

9.386438 

915 

10.613562 

42 

374452 

863 

987465 

51 

386987 

914 

613013 

18 

43 

374970 

862 

987434 

51 

387536 

913 

612464 

17 

44 

375487 

861 

987403 

52 

388084 

912 

611916 

16 

45 

376003 

860 

987372 

52 

388631 

911 

611369 

15 

46 

376519 

859 

987341 

52 

389178 

910 

610822 

14 

47 

377035 

858 

997310 

52 

389724 

909 

610276 

13 

48 

377549 

857 

987279 

52 

390270 

908 

609730 

12 

49 

378003 

856 

987248 

52 

390815 

907 

609185 

11 

50 

51 

378577 

854 

987217 
9.987186 

52 
52 

391360 

906 

608640 
10.608097 

10 
9 

9.379089 

853 

9.391903 

905 

52 

379601 

852 

987155 

52 

392447 

904 

607553 

8 

63 

380113 

851 

987124 

52 

392989 

903 

607011 

7 

54 

380624 

850 

987092 

52 

393531 

902 

606469 

6 

55 

381134 

849 

987061 

52 

394073 

901 

.  605927 

5 

56 

381643 

848 

987030 

52 

394614 

900 

605386 

4 

67 

382152 

847 

986998 

52 

395154 

899 

604846 

3 

58 

382661 

846 

986967 

52 

395694 

898 

604306 

2 

59 

383168 

845 

986936 

52 

396233 

897 

603767 

1 

60 

383675 

844 

986904 

52 

396771 

896 

603229 

0 

^ 

Cosine   1 

1 

Sine   1    1 

Cotang.  1 

Tang.  |M.  1 

76  Degrees. 


32 

(14  Degrees.;  a 

TABLE  OF  LOGARITHMIC 

pn 

Sine   1 

D. 

Cosine   |  D. 

Tang.   1 

D.   1 

Cotang.   1   1 

0 

9.383675 

844 

9.986904 

52 

9.396771 

896 

10.603229 

60 

1 

384182 

843 

986873 

53 

397309 

896 

602691 

59 

2 

384687 

842 

986841 

53 

397846 

895 

602154 

58 

3 

385192 

841 

986809 

53 

398383 

894 

601617 

57 

4 

385697 

840 

986778 

53 

398919 

893 

601081 

66 

5 

386201 

839 

986746 

53 

399455 

892 

600545 

55 

6 

386704 

838 

986714 

53 

399990 

891 

600010 

54 

7 

387207 

83T 

986683 

53 

400524 

890 

599476 

53 

8 

387709 

836 

986651 

53 

401058 

889 

598942 

52 

9 

388210 

835 

986619 

53 

401591 

888 

598409 

51 

10 
11 

388711 

834 

986587 

53 
53 

402124 

887 

597876 

50 
49 

9.389211 

833 

9.986555 

9.402656 

886 

10.597344 

12 

389711 

832 

986523 

53 

403187 

885 

696813 

48 

13 

390210 

831 

986491 

53 

403718 

884 

596282 

47 

14 

390708 

830 

986459 

53 

404249 

883 

695751 

46 

15 

391206 

828 

986427 

53 

404778 

882 

595222 

45 

16 

391703 

827 

986395 

53 

405308 

881 

694692 

44 

17 

392199 

826 

986363 

54 

405836 

880 

594164 

43 

18 

392695 

825 

986331 

54 

406364 

879 

593636 

42 

19 

393191 

824 

986299 

54 

406892 

878 

593108 

41 

20 
21 

393685 

823 

986266 
9.986234 

54 
54 

407419 

877 

592581 

40 
39 

9.394179 

822 

9.407945 

876 

10.592055 

22 

394673 

821 

986202 

54 

408471 

875 

591529 

38 

23 

395166 

820 

986169 

54 

408997 

874 

691003 

37 

24 

395658 

819 

986137 

54 

409521 

874 

690479 

36 

25 

396150 

816 

986104 

54 

410045 

873 

689955 

35 

26 

396641 

817 

986072 

54 

410569 

872 

589431 

34 

27 

3971-32 

817 

986039 

54 

411092 

871 

588908 

33 

28 

397621 

816 

986007 

54 

411615 

870 

588385 

32 

29 

398111 

815 

985974 

54 

412137 

869 

587863 

31 

30 
31 

398600 

814 

985942 
9.985909 

54 
55 

412658 

868 

687342 

30 
29 

9.399088 

813 

9.413179 

867 

10.586821 

32 

399575 

812 

-  985876 

55 

413699 

866 

586301 

28 

33 

400062 

811 

985843 

55 

414219 

865 

585781 

27 

34 

400549 

810 

985811 

55 

414738 

864 

585262 

26 

35 

401035 

809 

985778 

55 

41.5257 

864 

584743 

26 

36 

401520 

808 

.  985745 

55 

415775 

863 

684225 

24 

37 

402005 

807 

985712 

55 

416293 

862 

5837-07 

23 

38 

402489 

806 

985679 

55 

416810 

861 

583190 

22 

39 

402972 

805 

985646 

55 

417326 

860 

682674 

21 

40 
41 

403455 

804 

985613 

55 
55 

417842 

859 

582158 

20 
19 

9.403938 

803 

9.985580 

9.418358 

858 

10.581642 

42 

404420 

802 

985547 

55 

418873 

857 

581127 

18 

43 

404901 

801 

985514 

55 

419387 

856 

580613 

17 

44 

405382 

800 

985480 

55 

419901 

855 

580099 

16 

45 

405862 

799 

985447 

55 

420415 

855 

579585 

15 

46 

406341 

798 

985414 

56 

420927 

854 

579073 

14 

47 

406820 

.797 

985380 

56 

421440 

853 

678560 

13 

48 

407299 

796 

985347 

56 

421952 

852 

678048 

12 

49 

407777 

795 

985314 

56 

422463 

851 

677537 

11 

50 
51 

408254 

794 

985280 
9.985247 

56 
56 

422974 

850 

677026 

10 
9 

9.408731 

794 

9.423484 

849 

10.576516 

52 

409207 

793 

985213 

56 

423993 

848 

576007 

8 

53 

409682 

792 

985180 

56 

424503 

848 

675497 

7 

54 

410157 

791 

985146 

56 

425011 

847 

674989 

6 

55 

410632 

790 

985113 

56 

425519 

846 

674481 

5 

56 

411106 

789 

985079 

56 

426027 

845 

673973 

4 

57 

411579 

788 

985045 

56 

426534 

844 

573466 

3 

58 

412052 

787 

985011 

56 

427041 

843 

572959 

2 

59 

412524 

786 

984978 

56 

427547 

843 

672453 

1 

^ 

412996 

785 

984944!  56 

428052 

842 

571948 

0 

^ 

Cosine 

Sine   1 

j  Uoiang. 

1   Tang  =,M.| 

75  Degrees. 


SINES  AND  TANGENTS.   (15  Degrees.) 

33 

M. 

1    Sine 

1   D. 

1  Cosine   1  D. 

1   Tang. 

1   D. 

1   Cotang.  1    1 

^ 

9.412996 

785 

9.984944 

57 

9.428052 

842 

10.571948 

60 

1 

413467 

784 

984910 

57 

428557 

841 

571443 

59 

2 

413938 

783 

984876 

57 

429062 

840 

670938 

58 

3 

414408 

783 

984842 

57 

429566 

839 

670434 

57 

4 

414878 

782 

984808 

57 

430070 

838 

569930 

56 

6 

415347 

781 

984774 

57 

430573 

838 

569427 

55 

6 

415815 

780 

984740 

57 

431075 

837 

668926 

54 

7 

416283 

779 

994706 

57 

431677 

836 

568423 

53 

8 

416751 

778 

984672 

57 

432079 

835 

667921 

62 

9 

417217 

777 

984637 

57 

432680 

834 

567420 

51 

10 
11 

417684 
9. 41-8150 

770 

984603 
9.984569 

57 
57 

433080 

833 

566920 

50 
49 

775 

9.433580 

832 

10.. 566420 

12 

418615 

774 

984535 

57 

434080 

832 

565920 

48 

13 

419079 

773 

984500 

57 

434579 

831 

565421 

47 

14 

419544 

773 

984466 

57 

435078 

830 

564922 

46 

16 

420007 

772 

984432 

58 

435576 

829 

564424 

45 

16 

420470 

771 

984397 

58 

436073 

828 

563927 

44 

17 

420933 

770 

984363 

58 

436570 

828 

563430 

43 

18 

421395 

769 

984328 

58 

437067 

827 

662933 

42 

19 

421857 

768 

984294 

58 

437563 

826 

662437 

41 

20 
21 

422318 
9  422778 

767 

984259 

58 
58 

438059 
9.438554 

825 

824 

661941 

40 
39 

767 

9.984224 

10.561446 

22 

423238 

766 

984190 

58 

439048 

823 

560952 

38 

23 

423697 

765 

984155 

58 

439543 

823 

560457 

37 

24 

424156 

764 

984120 

58 

440036 

822 

569964 

36 

25 

424615 

763 

984085 

58 

440529 

821 

559471 

36 

26 

425073 

762 

984050 

58 

44102a 

820 

568978 

34 

27 

425530 

761 

984015 

58 

441514 

819 

558486 

33 

28 

425987 

760 

983981 

58 

442006 

819 

557994 

32 

29 

426443 

760 

983946 

58 

442497 

818 

567503 

31 

30 
31 

426899 
9.427354 

759 

758 

983911 
9.983875 

58 
58 

442988 

817 

657012 

30 

29 

9.443479 

816 

10.656521 

32 

427809 

757 

983840 

59 

443968 

816 

566032 

28 

33 

428263 

756 

983805 

59 

444458 

815 

555542 

27 

34 

428717 

755 

983770 

59 

444947 

814 

555053 

26 

35 

429170 

754 

983735 

59 

446435 

813 

554566 

25 

36 

429623 

753 

983700 

59 

445923 

812 

5.54077 

24 

37 

430075 

752 

983664 

59 

446411 

812 

653589 

23 

38 

430527 

752 

983629 

59 

446898 

811 

553102 

22 

39 

430978 

751 

983594 

59 

447384 

810 

652616 

21 

40 
41 

431429 

750 

983558 

59 
59 

447870 
9.448356 

809 

552130 

20 
19 

9.431879 

749 

9.983523 

809 

10.561644 

42 

432329 

749 

9S3487 

59 

448841 

808 

651159 

18 

43 

432778 

748 

983452 

69 

449326 

807 

650674 

17 

44 

433226 

747 

9834161 

59 

449810 

806 

550190 

16 

45 

433675 

746 

983381 

59 

450294 

806 

549706 

16 

46 

434122 

745 

983345 

59 

450777 

805 

549223 

14 

47 

434569 

744 

983309 

59 

451260 

804 

648740 

13 

48 

435016 

744 

983273 

60 

451743 

803 

548257 

12 

49 

435462 

743 

983238 

60 

452225 

802 

647775 

11 

60 
51 

435908 
9.436353 

742 

983202 

60. 
60 

452706 

802 

647294 
10.546813 

10 
9 

741 

9.983166 

9.453187 

801 

52 

436798 

740 

983130 

60 

463668 

800 

646332 

8 

53 

437242 

740 

983094 

60 

454148 

799 

645862 

7 

54 

437686 

739 

983058 

60 

454628 

799 

545372 

6 

55 

438129 

738 

98.3022 

60 

455107 

798 

544893 

6 

56 

438572 

737 

982986 

60 

456686 

797 

544414 

4 

57 

439014 

736 

982950 

60 

456064 

796 

543936 

3 

58 

439456 

736 

982914 

60 

456542 

796 

543468 

2 

59 

439897 

735 

982878 

60 

457019 

795 

542981 

1 

60 

440338 

734 

982842 

60 

457496 

794 

642504 

0 

1   1 

Cosine   j" 

1 

Sine   1   j 

Colang.  1 

1 

Tang.   1  M.  | 

74  Degrees. 

E 


34 

(16  Degrees.)  a 

TABLE  OP  LOGARITHMIC 

"m" 

Sine 

D.   1 

Cosine  | 

d:i 

Tang. 

D. 

Cotang.  1   1 

"01 

9.440338 

734 

9.982842 

60 

9.457496 

794 

10.542504 

60 

1 

440778 

733 

982805 

60 

457973 

793 

542027 

59 

2 

441218 

732 

982769 

61 

458449 

793 

541551 

68 

3 

441658 

731 

982733 

61 

458925 

792 

541075 

57 

4 

442096 

731 

982696 

61 

459400 

791 

540600 

56 

6 

442535 

730 

982660 

61 

459875 

790 

540125 

55 

6 

442973 

729 

982624 

61 

460349 

790 

539651 

54 

7 

443410 

728 

982587 

61 

460823 

789 

539177 

53 

8 

443847 

727 

982551 

61 

461297 

788 

538703 

52 

9 

444284 

727 

982514 

61 

461770 

788 

538230 

fl 

10 
11 

444720 

726 

982477 

61 
61 

462242 
9.462714 

787 
786 

537758 

50 
49 

9.445155 

725 

9.982441 

10.537286 

IS 

445590 

724 

982404 

61 

463186 

785 

536814 

48 

13 

446025 

723 

982367 

61 

463658 

785 

536342 

47 

14 

446459 

723 

982331 

61 

464129 

784 

535871 

46 

15 

446893 

722 

982294 

61 

464599 

783 

535401 

45 

16 

447326 

721 

982257 

61 

465069 

783 

634931 

44 

17 

447759 

720 

982220 

62 

465539 

782 

534461 

43 

18 

448191 

720 

982183 

62 

466008 

781 

633992 

42 

19 

448623 

719 

982146 

62 

466476 

780 

533524 

41 

20 
21 

449054 

718 

982109 

62 
62 

466945 

780 

533055 
10.532587 

40 
39 

9.449485 

717 

9.982072 

9.467413 

779 

22 

449915 

716 

982035 

62 

467880 

778 

532120 

38 

T3 

450346 

716 

981998 

62 

468347 

778 

531'653 

37 

24 

450775 

715 

981961 

62 

468814 

777 

531186 

36 

25 

451204 

714 

981924 

62 

469280 

776 

530720 

35 

26 

451632 

713 

981886 

6-^ 

409r4tJ 

775 

530254 

34 

27 

462060 

713 

981849 

62 

470211 

775 

529789 

33 

28 

452488 

712 

981812 

62 

470676 

774 

529324 

32 

29 

452915 

711 

981774 

62 

471141 

773 

528859 

31 

30 
31 

453342 

710 

981737 
9.981699 

62 
63 

471605 
9.472068 

773 

528395 

30 
29 

9.453768 

710 

772 

10.527932 

32 

454194 

709 

981662 

63 

472532 

771 

527468 

28 

33 

454619 

708 

981625 

63 

472995 

771 

527005 

27 

34 

455044 

707 

981587 

63 

473457 

770 

526543 

26 

35 

455469 

707 

981549 

63 

473919 

769 

626081 

26 

36 

455893 

706 

981512 

63 

474381 

769 

525619 

24 

37 

456316 

705 

981474 

63 

474842 

768 

525158 

23 

38 

456739 

704 

981436 

63 

475303 

767 

524697 

22 

39 

457162 

704 

981399 

63 

475763 

767 

524237 

21 

40 

457584 

703 

981361 

63 

476223 

766 

623777 

20 

41 

9.458006 

702 

9.981323 

63 

9.476683 

765 

10.523317 

19 

42 

458427 

701 

981285 

63 

477142 

765 

622858 

18 

43 

458848 

701 

981247 

63 

477601 

764 

522399 

17 

44 

459268 

700 

981209 

63 

478059 

763 

521941 

16 

45 

459688 

699 

981171 

63 

478517 

763 

521483 

15 

46 

460108 

698 

981133 

64 

478975 

762 

521025 

14 

47 

460527 

698 

981095 

64 

479432 

761 

520568 

13 

48 

460946 

697 

981057 

64 

479889 

761 

620111 

12 

49 

461364 

696 

981019 

64 

480345 

760 

519655 

11 

50 
51 

461782 

695 

980981 

64 
64 

480801 

759 

519199 
10.518743 

10 
9 

9.462199 

695 

9.980942 

9.481257 

759 

52 

462616 

694 

980904 

64 

481712 

758 

618288 

8 

53 

463032 

693 

980866 

64 

482167 

757 

517833 

7 

54 

463448 

693 

980827 

64 

482621 

757 

517379 

6 

55 

463864 

692 

980789 

64 

483075 

756 

516925 

5 

56 

464279 

691 

980750 

64 

483529 

755 

516471 

4 

57 

464694 

690 

980712 

64 

483982 

755 

616018 

3 

58 

465108 

690 

980673 

64 

484435 

754 

515566 

2 

59 

465522 

689 

980635 

64 

484887 

753 

615113 

1 

60 

465935 

688 

980596 

64 

485339 

753 

514661 

0 

Cosine 

1 

1   Sine   1 

1  Colang. 

1 

Tang.   1  M.  | 

73  Degrees. 


SINES  AND  TANGENTS.  (^17  Degrees.) 

35 

M. 

Sine 

D. 

Cosine   1  D. 

1   Tang. 

1   D. 

Cotang.   1   1 

0 

9.465935 

688 

9.980596 

64 

9.485339 

765 

10.614661 

60 

1 

466348 

688 

980558 

64 

485791 

•£62 

514209 

59 

2 

466761 

687 

980519 

65 

486242 

761 

613768 

68 

3 

467173 

686 

980480 

65 

486693 

751 

513307 

67 

4 

467585 

685 

980442 

65 

487143 

750 

612857 

56 

5 

467996 

685 

980403 

65 

487593 

749 

512407 

55 

6 

466407 

684 

980364 

65 

488043 

749 

511957 

64 

7 

468817 

683 

980325 

65 

488492 

748 

511508 

53 

8 

469227 

683 

980286 

66 

488941 

747 

511059 

52 

9 

469637 

682 

980247 

65 

489390 

747 

510610 

51 

10 
11 

470046 

681 

980208 
9.980169 

65 
65 

489838 
9.490286 

746 

610162 
10.609714 

60 
49 

9.470455 

680 

746 

12 

470863 

680 

980130 

65 

490733 

746 

509267 

48 

13 

471271 

679 

980091 

65 

491180 

744 

508820 

47 

14 

471679 

678 

980052 

65 

491627 

744 

608373 

46 

15 

472086 

678 

980012 

65 

492073 

743 

507927 

45 

16 

472492 

677 

979973 

65 

492519 

743 

607481 

44 

17 

472898 

676 

979934 

66 

492965 

742, 

507035 

43 

18 

473304 

676 

979895 

66 

493410 

741 

506590 

42 

19 

473710 

675 

979856 

66 

493854 

740 

506143 

41 

20 
21 

474115 

674 

979816 
9.979776 

66 
66 

494299 

740 

505701 
10.505257 

40 
39 

9.474519 

674 

9.494743 

740 

22 

474923 

673 

979737 

66 

496186 

739 

604814 

38 

23 

475327 

672 

979697 

66 

495630 

738 

604370 

37 

24 

475730 

672 

979658 

66 

496073 

737 

603927 

36 

25 

476133 

671 

979618 

66 

490515 

737 

503485 

35 

26 

476536 

670 

979579 

66 

496957 

736 

503043 

34 

27 

476938 

669 

979539 

66 

497399 

736 

502601 

33 

28 

477340 

669 

979499 

66 

497841 

735 

602159 

32 

29 

477741 

668 

979459 

66 

498282 

734 

501718 

31 

30 
31 

478142 

667 

979420 
9.979380 

66 
66 

498722 
9.499163 

734 
733 

501278 

30 
29 

9.478542 

667 

10.500837 

32 

478942 

666 

979340 

66 

499603 

733 

600397 

28 

33 

*'479342 

665 

979300 

67 

500042 

732 

499958 

27 

34 

479741 

665 

979260 

67 

500481 

731 

499519 

26 

35 

480140 

664 

979220 

67 

600920 

731 

499080 

26 

36 

480539 

663 

979180 

67 

501359 

730 

498641 

24 

37 

480937 

663 

979140 

67 

501797 

730 

498203 

23 

38 

481334 

662 

979100 

67 

502236 

729 

497765 

22 

39 

481731 

661 

979059 

67 

502672 

728 

497328 

21 

40 
41 

482128 

661 

979019 
9.978979 

67 
67 

503109 

728 

496891 
10.496454 

20 
19 

9.482525 

660 

9.503546 

727 

42 

482921 

659 

978939 

67 

503982 

727 

496018 

18 

43 

483316 

659 

978898 

67 

604418 

726 

495582 

17 

44 

483712 

658 

978868 

67 

504854 

725 

495146 

16 

45 

484107 

657 

978817 

67 

505289 

726 

494711 

15 

46 

484501 

657 

978777 

67 

505724 

724 

494276 

14 

47 

484895 

656 

978736 

67 

506159 

724 

493841 

13 

48 

485289 

655 

978696 

68 

506693 

723 

493407 

12 

49 

485682 

655 

978655 

68 

607027 

722 

492973 

11 

60 
51 

486075 
9.486467 

654 

978615 
9.978574 

68 
68 

507460 
9.507893 

722 
721 

492540 

10 
9 

653 

10.492107 

62 

486860 

653 

978533 

68 

608326 

721 

491674 

8 

53 

487251 

652 

978493 

68 

608759 

720 

491241 

7 

64 

487643 

651 

978452 

68 

509191 

719 

490809 

6 

65 

488034 

651 

978411 

68 

509622 

719 

490378 

6 

56 

488424 

650 

978370 

68 

510064 

718 

489946 

4 

57 

488814 

650 

978329 

68 

610485 

718 

489516 

3 

68 

489204 

649 

978288 

68 

510916 

717 

. 489084 

2 

59 

489693 

648 

978247 

68 

511346 

716 

488664 

1 

60 

489982 

648 

978206 

68 

611776 

716 

488224  1  0 

_l 

Cosine   1 

1 

Sine   1    1 

Cotang.  1 

'Jang.  1  M.~ 

72  Degrees. 


30 

(1 

B  Degr 

ees.)  A 

TABLE  OF  LOGARITHMIC 

"m" 

Sine 

D. 

1  Cosine   1  I). 

Tang. 

D. 

{   Cotang.      1 

0 

9.489982 

648 

9. 9782061 68 

9.511776 

716 

10.488224 

60 

1 

490371 

648 

978165 

68 

512206 

716 

487794 

59 

2 

490759 

647 

978124 

68 

512635 

715 

487365 

68 

3 

491147 

646 

978083 

69 

513064 

714 

486936 

57 

4 

491535 

646 

978042 

69 

513493 

714 

486507 

56 

5 

491922 

645 

978001 

69 

513921 

713 

486079 

56 

6 

492308 

644 

977959 

69 

614349 

713 

485651 

64 

7 

492695 

644 

977918 

69 

514777 

712 

485223 

53 

8 

493081 

643 

977877 

6y 

515204 

712 

484796 

52 

9 

493466 

642 

977835 

69 

515631 

711 

484369 

51 

10 

11 

493851 

642 

977794 

69 
69 

510057 
9.516484 

710 

483943 
10.483516 

50 
49 

9.494236 

641 

9.977752 

710 

12 

494621 

641 

977711 

69 

616910 

709 

483090 

48 

13 

495005 

640 

977669 

69 

517335 

709 

482665 

47 

14 

495388 

639 

977628 

69 

517761 

708 

482239 

46 

15 

495772 

639 

977586 

69 

518185 

708 

481815 

46 

16 

496154 

638 

977544 

70 

618610 

707 

481390 

44 

17 

496537 

637 

977503 

70 

519034 

706 

480966 

43 

18 

496919 

637 

977461 

70 

519458 

706 

480542 

42 

19 

497301 

636 

977419 

70 

619882 

705 

480118 

41 

20 
21 

497682 

636 

977377 

■70 
70 

620305 

705 

479695 
10.479272 

40 
39 

9.498064 

635 

9.977335 

9.520728 

704 

22 

498444 

634 

977293 

70 

621151 

703 

478849 

38 

23 

498825 

634 

977251 

70 

621673 

703 

478427 

37 

24 

499204 

633 

977209 

70 

621996 

703 

478005 

36 

25 

499584 

632 

977167 

70 

522417 

702 

477583 

35 

26 

499963 

632 

977125 

70 

522838 

702 

477162 

34 

27 

500342 

631 

977083 

70 

523259 

701 

476741 

33 

28 

.500721 

631 

977041 

70 

523680 

701 

476320 

32 

29 

501099 

630 

976999 

70 

524100 

700 

475900 

31 

30 
31 

501476 

629 

976957 

70 
70 

524520 

699 

475480 
10.475061 

30 
29 

9.501854 

629 

9.976914 

9.624939 

699 

32 

502231 

628 

976872 

71 

625359 

698 

474641 

28 

33 

502607 

628 

976830 

71 

625778 

698 

4742^2 

27 

34 

502984 

627 

976787 

71 

526197 

697 

473803 

26 

35 

503360 

026 

976745 

71 

626615 

697 

473385 

25 

36 

503735 

626 

976702 

71 

527033 

696 

472967 

24 

37 

504110 

625 

976660 

71 

527451 

696 

472549 

23 

38 

504485 

625 

976617 

71 

527868 

695 

472132 

22 

39 

504860 

624 

976574 

71 

528285 

695 

471715 

21 

40 

41 

605234 

623 

976532 

71 
71 

628702 

694 
693 

471298 

20 
19 

9.505608 

623 

y. 976489 

9.629119 

0.470881 

42 

505981 

622 

976446 

71 

629535 

693 

470465 

18 

43 

506354 

622 

976404 

71 

529950 

693 

470050 

17 

44 

506727 

621 

976361 

71 

530366 

692 

469634 

16 

45 

507099 

620 

976318 

71 

630781 

691 

469219 

16 

46 

507471 

620 

976275 

71 

531196 

691 

468804 

14 

47 

507843 

619 

976232 

72 

531611 

690 

468389 

13 

48 

508214 

619 

976189 

72 

632025 

690 

467975 

12 

49 

508585 

618 

976146 

72 

532439 

689 

467561 

11 

50 
61 

508956 
9.509326 

618 

976103 
9.976060 

72 

72 

532853 

689 

467147 
10.466734 

10 
9 

617 

9.533266 

688 

52 

509696 

616 

976017 

72 

533679 

688 

466321 

8 

53 

510065 

616 

976974 

72 

634092 

687 

465908 

7 

54 

510434 

615 

976930 

72 

534504 

687 

465496 

6 

55 

510803 

615 

975887 

72 

534916 

686 

465084 

6 

56 

511172 

614 

975844 

72 

535328 

686 

464672 

4 

57 

511540 

613 

975800 

72 

635739 

685 

464261 

3 

58 

511907 

613 

975757 

72 

636150 

685 

463850 

2 

59 

612275 

612 

975714 

72 

536561 

684 

463439 

1 

60 

512642 

612 

975670 

72 

536972 

684 

463028 

0 

~ 

Cosine 

Sine   1 

Cotang. 

Tang.    |M.| 

71  Degrees. 


SINES  AND  TANGENTS.  (19  Degrees.) 

37 

M. 

1   Siae 

1   D. 

1  Cosine  |  D. 

1   Tang. 

1   D. 

1   Cotang.   1   1 

~o" 

9.512642 

612 

9.975670 

73 

9.536972 

684 

10.463028 

60 

1 

513009 

611 

975627 

73 

637382 

683 

462618 

59 

2 

513375 

611. 

975583 

73 

537792 

683 

462208 

58 

3 

513741 

610 

975539 

73 

538202 

682 

461798 

67 

4 

514107 

609 

975496 

73 

538611 

682 

461389 

66 

5 

514472 

609 

975452 

73 

539020 

681 

460980 

56 

6 

514837 

608 

975408 

73 

539429 

681 

460571 

54 

7 

515202 

608 

9r5365 

73 

539837 

680 

460163 

53 

8 

515566 

607 

975321 

73 

.540245 

680 

459755 

62 

9 

516930 

607 

975277 

73 

540653 

679 

459347 

61 

10 
11 

516294 

606 
605 

975233 

73 
73 

541061 

679 

458939 
10.458532 

50 
49 

9.516657 

9.975189 

9.541468 

678 

12 

517020 

605 

975145 

73 

541875 

678 

458125 

48 

13 

517382 

604 

975101 

73 

.542281 

677 

457719 

47 

14 

517745 

604 

975057 

73 

542688 

677 

457312 

46 

15 

618107 

603 

975013 

73 

543094 

676 

456906 

45 

16 

518468 

603 

974969 

74 

543499 

676 

456501 

44 

17 

518829 

602 

974925 

74 

543906 

675 

456095 

43 

18 

619190 

601 

974880 

74 

544310 

675 

455690 

42 

19 

519551 

601 

974836 

74 

544715 

674 

455285 

41 

20 
21 

519911 

600 

974792 
9.974748 

74 
74 

545119 

674 
673 

454881 
10.464476 

40 
39 

9.520271 

600 

9.545524 

22 

520631 

699 

974703 

74 

545928 

673 

454072 

38 

23 

520990 

599 

974659 

74 

546331 

672 

453669 

37 

24 

65J1349 

598 

974614 

74 

546735 

672 

453265 

36 

25 

5^1707 

598 

974570 

74 

547138 

671 

452862 

35 

26 

522066 

597 

974525 

74 

547540 

671 

452460 

34 

27 

522424 

596 

974481 

74 

547943 

670 

452057 

33 

28 

522781 

696 

974436 

74 

548345 

670 

451655 

32^ 

29 

523138 

595 

974391 

74 

548747 

669 

451253 

31 

30 
31 

523495 

595 

974347 
9.974302 

76 
75 

549149 
9.549560 

669 

450851 

30 
29 

9.523852 

594 

668 

10.4504.50 

32 

-  524208 

694 

974257 

75 

^549951 

668 

450049 

28 

33 

524564 

593 

974212 

75 

550352 

667 

449648 

27 

34 

^24920 

593 

974167 

75 

650752 

667 

449248 

26 

35 

62B275 

592 

974122 

75 

651152 

666 

448848 

25 

36 

6256.30 

591 

974077 

75 

551652 

666 

448448 

24 

37 

525984 

591 

974032 

75 

651952 

665 

448048 

23  i 

38 

526339 

590 

973987 

75 

55235-1 

665 

447649 

22 

39 

526693 

590 

973942 

75 

552750 

665 

447250 

21 

40 
41 

527046 

589 

973897 

75 

75 

553149 
9.653548 

664 
664 

446851 

20 
19 

9.527400 

589 

9.973852 

10.446452 

42 

527753 

588 

973807 

75 

553946 

663 

446054 

18 

43 

528105 

588 

973761 

75 

654344 

663 

445056 

17 

44 

528458 

587 

973716 

76 

554741 

662 

445259 

16 

45 

528810 

587 

973671 

76 

555139 

662 

444801 

15 

46 

629161 

586 

973625 

76' 

555536 

681 

444464 

14 

47 

529513 

586 

973580 

76 

555933 

661 

444067 

13 

48 

529864 

585 

973535 

76 

556329 

660 

443671 

12 

49 

.^^302 15 

585 

973489 

76 

656725 

660 

443275 

11 

50 
51 

.'530565 

584 

973444 

76 
76 

557121 

659 

442879 

10 
9 

9.530915 

684 

9.973398 

9.557517 

659 

10.442483 

52 

531265 

583 

973352 

76 

557913 

659 

442087 

8 

63 

531614 

582 

973307 

76 

558308 

658 

441692 

7 

54 

531963 

582 

973261 

76 

558702 

658 

441298 

6 

55 

532312 

581 

973215 

76 

559097 

657 

440903 

6 

56 

632661 

581 

973169 

76 

559491 

657 

440609 

4 

57 

633009 

580 

973124 

76 

559885 

656 

440115 

3 

58 

533357 

580 

973078 

76 

560279 

656 

439721 

2 

59 

533704 

579 

973032 

77 

560673 

655 

439327 

1 

60 

534052 

578 

972986 

77 

561066 

655 

438934 

0 

n 

Cosine 

Sine   1 

Cotang. 

1 

Tang.    1  M.  | 

Dd 


70  Degrees. 


'M 


38 

(20  Degrees.)  a 

TABLE  OF  LOGARITHMIC 

M. 

1   Sine 

D. 

1  Cosine   |  D. 

Tang. 

1   D. 

Cotang.   1   1 

"o" 

9.534052 

578 

9.9729861  77 

9.561066 

655 

10.438934 

60 

1 

534399 

577 

972940 

77 

561459 

654 

438541 

59 

2 

534745 

677 

972894 

77 

561851 

654 

438149 

58 

3 

535092 

577 

972848 

77 

562244 

653 

437756 

57 

4 

535438 

676 

972802 

77 

662636 

653 

437364 

56 

5 

535783 

576 

972755 

77 

663028 

653 

436972 

55 

6 

536129 

575 

972709 

77 

563419 

652 

436681 

54 

7 

536474 

574 

972663 

77 

563811 

652 

436189 

53 

<  8 

536818 

574 

972617 

77 

564202 

651 

435798 

62 

9 

537163 

573 

972570 

77 

564592 

651 

435408 

51 

10 
11 

537507 
9.537851 

573 

972524 
9.972478 

77 
77 

564983 

650 

435017 

50 
49 

572 

9.565373 

650 

10.434627 

12 

538194 

672 

972431 

78 

565763 

649 

434237 

48 

13 

538538 

571 

972385 

78 

566153 

649 

433847 

47 

14 

638880 

571 

972338 

78 

566542 

649 

433458 

46- 

15 

53^9223 

570 

972291 

78 

566932 

648 

433068 

45 

16 

539565 

570 

972245 

78 

567320 

648 

432680 

44 

17 

639907 

569 

972198 

78 

567709 

647 

432291 

43 

18 

540249 

669 

972151 

78 

568098 

647 

431902 

42 

19 

540590 

568 

972105 

79 

568486 

646 

431514 

41 

20 
21 

540931 
9.541272 

.568 

972058 

78 
78 

568873 

646 

431127 

40 
39 

567 

9.972011 

9.569261 

64o 

10.430739 

22 

541613 

567 

971964 

78 

569648 

645 

430352 

38 

23 

541953 

566 

971917 

78 

570035 

645 

429965 

37 

24 

542293 

566 

971870 

78 

570422 

644 

42957& 

36 

25 

542632 

565 

971823 

78 

570809 

644 

429191 

35 

26 

542971 

565 

971776 

78 

671195 

643 

428805 

34 

27 

543310 

564 

971729 

79 

571581 

643 

428419 

33 

28 

543649 

564 

971682 

79 

571967 

642 

428033 

32 

29 

543987 

563 

971635 

79 

572352 

642 

427648 

31 

30 

544325 

563 

971588 

79 

572738 

642 

427262 

30 

31 

9.544663 

562 

9.971540 

79 

9.573123 

641 

10.426877 

29 

32 

545000 

562 

971493 

79 

573507 

641 

426493 

28 

33 

545338 

561 

971446 

79 

573892 

640 

426108 

27 

34 

545674 

561 

971398 

79 

574276 

640 

425724 

26 

35 

546011 

560 

971.%1 

79 

574660 

630 

425340 

25 

36 

546347 

660 

971303 

79 

576044 

639 

424956 

24 

37 

546683 

559 

&71256 

79 

575427 

639 

424573 

23 

38 

547019 

559 

971208 

79 

575810 

638 

424190 

22 

39 

547354 

558 

971161 

79 

576193 

638 

423807 

21 

40 
41 

547689 

658 

971113 
9.971066 

79 

80 

576576 

637 

423424 

20 
19 

9.548024 

557 

9.576958 

637 

10.423041 

42 

548359 

557 

971018 

80 

577341 

636 

422659 

18 

43 

548693 

656 

970970 

80 

577723 

636 

422277 

17 

44 

549027 

556 

970922 

80 

578104 

636 

421896 

16 

45 

549360 

655 

970874 

80 

578486 

635 

421514 

15 

46 

549693 

655 

970827 

80 

578867 

635 

421133 

14 

47 

550026 

554 

970779 

80 

579248 

634 

420752 

13 

48 

550359 

554 

970731 

80 

579629 

634 

420371 

12 

49 

550692 

553 

970683 

80 

580009 

634 

419991 

11 

50 
61 

551024 
9.551356 

553 

970635 
9.970586 

80 
80 

580389 

633 

419611 
10.419231 

10 
9 

-  552 

9.580769 

633 

52 

551687 

652 

970538 

80 

581149 

632 

418851 

8 

63 

552018 

552 

970490 

80 

581528 

632 

418472 

7 

54 

552349 

551 

970442 

80 

581907 

632 

418093 

6 

55 

552680 

551 

970394 

80 

582286 

631 

417714 

5 

56 

553010 

560 

970345 

81 

582665 

631 

417335 

4 

57 

5.'>3341 

550 

970297 

81 

583043 

630 

416957 

3 

58 

553670 

649 

970249 

81 

683422 

630 

416578 

2 

59 

554000 

649 

970200 

81 

583800 

629 

416200 

1 

60 

654329 

548 

970152 

81 

584177 

629 

415823 

0 



1  Cosine 

1   Sine   1 

Ootaug. 

1   Tang. 

■mT 

Degrees. 


•^ 


•LY- 


SINES AND  TANGENTS.     (21  Degrees.) 


39 


T\ 

Sine   1 

D.   1 

Cosine  |  D.  \ 

Tang.   1 

D.   1 

Cotang.  1   1 

: 

0 

9.554329 

548 

9,970152 

81 

9.584177 

629 

10.415823 

60 

1 

554658 

548 

970103 

81 

584555 

629 

415445 

59 

2 

554987 

647 

970055 

81' 

584932 

628 

415068 

58 

3 

555315 

547 

970006 

81 

585309 

628 

414691 

57 

4 

555643 

546 

969957 

81 

585686 

627 

414314 

56 

6 

555971 

546 

969909 

81 

586062 

627 

413938 

55 

6 

556299 

545 

969860 

81 

586439 

627 

413.561 

54 

7 

656626 

545 

969811 

81 

586815 

626 

413185 

53 

8 

556953 

644 

969762 

8i 

587190 

626 

412810 

52 

9 

557280 

544 

969714 

81 

587566 

625 

41-2434 

51 

10 
11 

557606 

543 

969665 

81 

82 

587941 

625 

625 

•  412059 
10.411684 

50. 
49 

9.557932 

543 

9.969616 

9.588316. 

12 

558258 

543 

969667 

82 

588691 

624 

411309 

48 

13 

658583 

542 

969518 

82 

589066 

624 

410934 

47 

14 

558909 

642 

969469 

82 

589440 

623 

410560 

46 

15 

569234 

641 

969420 

82 

589814 

623 

410186 

45 

t 

16 

569568 

641 

969370 

82 

590188 

623 

409812 

44 

17 

56988a 

640 

969321 

82 

690562 

622 

409438 

43 

18 

660207 

540 

969272 

82 

590935 

622 

409065 

42 

19 

660531 

539 

969223 

82 

591308 

622 

408692 

41 

20 

21 

660865 

639 

969173 

82 

82 

591681 

621 

408319 

40! 
39 

9.561178 

638 

9.969124 

9.592054 

621 

10.407946 

22 

661501 

638 

969075 

82 

592426 

620 

407574 

38  > 

23 

661824 

637 

969025 

82 

592798 

620 

407202 

37' 

24 

662146 

63? 

968976 

82 

693170 

619 

406829 

36^ 

25 

662468 

536 

968926 

83 

593542 

619 

406458 

35 

26 

662790 

536 

9688.77 

88 

693914 

618 

406086 

34 

27 

663112 

536 

968827 

83 

594285 

618 

405715 

33 

28 

663433 

535 

968777 

83 

594656 

618 

405344 

32- 

29 

663756 

535 

968728 

83 

595027 

617 

404973 

31 

30 
31 

664075 

534 

968678 
9.968628 

83 
83 

695398 

617 

404602 

30 

29 

9,564396 

534 

9.595768 

617 

10.404232 

32 

664716 

533 

968578 

83 

596138 

616 

403862 

28 

33 

665036 

.633 

968528 

83 

596508 

616 

403492 

27. 

34 

605356 

'532 

968479 

83 

596878 

616 

403122 

26^ 

35 

565676 

632 

968429 

83 

597247 

615 

402753 

25 

36 

665996 

631 

968379 

83 

597616 

615 

402384 

24 

37 

666314 

631 

968329 

83 

697985 

615 

402015 

23 

38 

566632 

531 

968278 

83 

5(98354 

614 

401646 

22^ 

39 

666951 

630 

968228 

84 

598722 

614 

401278 

21 

40 
41 

567269 

630 

968178 
9.908128 

84 
«4 

599091 

613 

400909 

20 
19 

9.567587 

629 

9.599459 

613 

10.400541 

42 

667904 

529 

968078 

84 

599827 

613 

400173 

18 

43 

568222 

528 

968027 

84 

600194 

612 

399806 

17 

44 

668539 

528 

967977 

84 

6005612 

612 

i399438 

16 

45 

568866 

'.528 

!967927 

84 

600929 

611 

399071 

15 

46 

5691:72 

,527 

-967870 

84 

601296 

611 

398704 

14 

47 

669488 

,527 

967826 

84 

601602 

611 

•398338 

13 

48 

569804 

526 

96777S 

84 

60202*] 

610 

397971 

12 

49 

670120 

526 

967725 

84 

602395 

610 

397605 

11 

50 
51 

57043S 

625 

967674 
9.967624 

84 
84 

60^61 
9.603127 

610 
609 

397239 
10.396873 

10 
9 

9.570751 

525 

52 

571066 

524 

96757? 

f  84 

603492 

609 

396507 

8 

53 

571361 

524 

967525 

85 

60385€ 

609 

396142 

7 

54 

571 69£ 

)   623 

96747] 

85 

60422S 

008 

395777 

6 

55 

57200t 

>   623 

•96742] 

85 

60458€ 

608 

395412 

5 

56 

57232C 

i      623 

96737( 

)  85 

60495S 

607 

395047 

4 

57 

672636 

)   622 

9673  U 

85 

605317 

607 

394683 

3 

58 

67295( 

)   622 

•967^6^ 

1  85 

605685 

607 

394318 

2 

59 

673261 

J   621 

96721" 

r   85 

60604t 

606 

393954 

1 

60 

67357f 

)   621 

9671061 85 

60641C 

606 

393590 

0 

1  Cosine 

1 

1   Sine   1 

1  Cotang. 

1 

1    Tang.   1  M. 

fi8  Degrees- 


40 


(22  Degrees.)     a  table  op  logarithmic 


M. 

I   Sine 

1   D. 

Cosine   |  D. 

1   Tang. 

D. 

Cotang. 

' 

^ 

9.573575 

521 

9.967166 

85 

9.606410 

606 

10.393590 

"6(f 

1 

573888 

520 

967115 

85 

606773 

606 

393227 

59 

2 

574200 

520 

967064 

85 

607137 

605 

392863 

58 

3 

574512 

519 

967013 

85 

607500 

605 

392500 

57 

4 

574824 

519 

966961 

85 

607863 

604 

392137 

56 

5 

675136 

619 

966910 

85 

608225 

604 

391775 

55 

6 

575447 

518 

966859 

85 

608588 

604 

391412 

54 

7 

575758 

518 

966808 

85 

608950 

603 

391050 

53 

8 

676069 

517 

966756 

86 

609312 

603 

390688 

52 

9 

576379 

517 

966705 

86 

609674 

603 

390326 

51 

10 
11 

576689 

516 

966663 
9.966602 

86 
86 

610036 
9.610397 

602 

389964 
10.389603 

60 
49 

9.576999 

616 

602 

12 

577309 

616 

966550 

86 

610769 

602 

389241 

48 

13 

577618 

515 

966499 

86 

611120 

601 

388880 

47 

14 

577927 

516 

966447 

86 

611480 

601 

388520 

46 

15 

578236 

514 

966395 

86 

611841 

601 

388169 

45 

16 

578545 

514 

966344 

86 

612201 

600 

387799 

44 

17 

578853 

513 

966292 

86 

612561 

600 

387439 

43 

18 

579162 

513 

966240 

86 

612921 

600 

387079 

42 

19 

579470 

513 

966188 

86 

613281 

599 

386719 

41 

20 
21 

579777 

512 

966136 
9  966085 

86 
87 

613641 

599 

386359 

40 
39 

9.680085 

612 

9.614000 

598 

10.386000 

22 

580392 

511 

966033 

87 

614359 

598 

385641 

38 

23 

680699 

511 

965981 

87 

614718 

598 

385282 

37 

24 

581005 

511 

965928 

87 

616077 

597 

384923 

36 

25 

581312 

610 

965876 

87 

615435 

597 

384565 

35 

26 

681618 

510 

965824 

87 

615793 

597 

384207 

34 

27 

581924 

509 

965772 

87 

616151 

596 

383849 

33 

28 

582229 

509 

965720 

87 

616509 

696 

383491 

32 

29 

582535 

509 

965668 

87 

616867 

596 

383133 

31 

30 
31 

582840 

508 

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1 

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D4 


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42 

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TABLE  OF  LOGABITHMIC 

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0 

9.609313 

473 

9.960730 

94 

9.648583 

566 

10.351417 

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4 

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565 

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6 

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94 

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7 

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94 

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564 

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53 

8 

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470 

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94 

651297 

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52 

9 

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94 

651636 

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51 

10 
11 

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94 
95 

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563 

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50 
49 

9.612421 

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9.960109 

9.652312 

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10.347688 

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13 

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18 

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19 

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41 

20 

21 

614944 

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9.959539 

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10.344316 

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38 

23 

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24 

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25 

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30 
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22 

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97 

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21 

40 

41 

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97 

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9.662376 

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101 

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6,34778 

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101 
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10.317937 

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101 

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43 

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101 

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44 

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31.5999 

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49 

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31.5354 

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50 

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10 

51 

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9.954213 

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9.686290 

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52 

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102 

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53 

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54 

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313745 

6 

55 

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56 

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4 

57 

641064 

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58 

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687540 

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59 

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953722 

103 

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44 

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TABLE  OF  LOGARITHMIC 

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0 

9.641842 

431 

9.953660 

103 

9.688182 

534 

10.311818 

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1 

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431 

953599 

103 

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311498 

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2 

642360 

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953537 

103 

688823 

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311177 

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3 

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953475 

103 

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533 

310857 

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4 

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103 

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310537 

56 

5 

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953352 

103 

689783 

533 

310217 

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6 

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953290 

103 

690103 

533 

309897 

54 

7 

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429 

953228 

103 

690423 

533 

309577 

53 

8 

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429 

953166 

103 

690742 

532 

309258 

52 

9 

644165 

429 

953104 

103 

691062 

532 

308938 

51 

10 
11 

644423 

428 

953042 

103 

104 

691381 

532 

308619 
10.308300 

50 
49 

9.644680 

428 

9.952980 

9.691700 

531 

12 

644936 

428 

952918 

104 

692019 

531 

307981 

48 

13 

645193 

427 

952855 

104 

692338 

531 

307662 

47 

14 

645450 

427 

952793 

104 

692656 

531 

307344 

46 

15 

645706 

427 

952731 

104 

692975 

531 

307025 

45 

16 

645962 

426 

952669 

104 

693293 

530 

306707 

44 

17 

646218 

426 

952606 

104 

693612 

530 

306388 

43 

18 

646474 

426 

952544 

104 

693930 

530 

306070 

42 

19 

646729 

425 

952481 

104 

694248 

530 

305752 

41 

20 
21 

646984 

425 

952419 
9.952356 

104 

694566 

529 

305434 

40 
39 

9.647240 

425 

9.694883 

529 

10.3'05117 

22 

647494 

424 

952294 

104 

695201 

529 

304799 

38 

23 

647749 

424 

952231 

104 

695518 

529 

304482 

37 

24 

648004 

424 

952168 

105 

695836 

529 

304164 

36 

25 

648258 

424 

952106 

105 

696153 

528 

303847 

35 

36 

648512 

423 

952043 

105 

696470 

528 

303530 

34 

27 

648766 

423 

951980 

105 

696787 

528 

303213 

33 

28 

649020 

423 

951917 

105 

697103 

528 

302897 

32 

29 

649274 

422 

951854 

105 

697420 

527 

302580 

31 

30 
31 

649527 

422 

951791 

105 
105 

697736 

527 

302264 

30 
29 

9.649781 

422 

9.951728 

07698053 

527 

10.301947 

32 

650034 

422 

951665 

105 

698369 

527 

.301631 

28 

33 

650287 

421 

951602 

105 

698685 

526 

301315 

27 

34 

650539 

421 

951539 

105 

699001 

526 

300999 

26 

35 

650792 

421 

951476 

105 

699316 

526 

300684 

25 

36 

651044 

420 

951412 

105 

699632 

526 

300368 

24 

37 

651297 

420 

951349 

106 

699947 

526 

300053 

23 

38 

651549 

420 

951286 

106 

700263 

525 

299737 

22 

39 

651800 

419 

951222 

106 

700578 

525 

299422 

21 

40 
41 

652052 
9.652304 

419 

951159 

106 
106 

700893 

525 

299107 
10.298792 

20 
19 

419 

9.951096 

9.701208 

524 
524 

42 

652555 

418 

951032 

106 

701523 

298477 

18 

43 

652806 

418 

9509G8 

106 

701837 

524 

298163 

17 

44 

653057 

418 

950905 

106 

702152 

524 

297848 

16 

45 

653308 

418 

950841 

106 

702466 

524 

297534 

15 

46 

653558 

417- 

950778 

106 

702780 

523 

297220 

14 

47 

653808 

417 

950714 

106 

703095 

523 

296905 

13 

48 

654059 

417 

950650 

106 

703409 

523 

296591 

12 

49 

654309 

416 

950586 

106 

703723 

523 

296277 

11 

50 
51 

654558 

416 

950522 
9.950458 

107 
107 

704036 

522 

295964 
10.295650 

10 
9 

9.654808 

416 

9.704350 

522 

52 

655058 

416 

950394 

107 

704663 

522 

295337 

8 

53 

655307 

415 

950330 

107 

704977 

522 

295023 

7 

.54 

655556 

415 

950266 

107 

705290 

522 

294710 

6 

55 

655805 

415 

950202 

107 

705603 

521 

294397 

5 

56 

656054 

414 

950138 

107 

705916 

521 

294084 

4 

57 

656302 

414 

950074 

107 

706228 

521 

293772 

3 

58 

656551 

414 

950010 

107 

706541 

521 

293459 

2 

59 

656799 

413 

940945 

107 

706854 

521 

293146 

1 

60 

657047 

413 

949881 

107 

707166 

520 

292834 

0 

Cosine 



Sine   1 

Cotan?. 



1   Tang.    1  M.  | 

G3  Degrees. 


SINES  AND  TANGENTS.  (27  Degrees.) 

45 

M. 

Sine 

D. 

Cosine  |  D. 

Tang. 

1   D. 

1  Cotang.   1   1 

"o" 

{>.  657047 

413 

9.949881 

107 

9.707166 

520 

10.292834 

60 

1 

657295 

413 

949816 

107 

707478 

620 

292622 

59 

2 

657542 

412 

949752 

107 

707790 

620 

292210 

58 

3 

657790 

412 

949688 

108 

708102 

520 

291898 

57 

4 

658037 

412 

949623 

108 

708414 

519 

291.586 

56 

5 

658284 

412 

949558 

108 

708726 

519 

291274 

56 

6 

658531 

411 

949494 

108 

709037 

619 

290963 

54 

7 

658778 

411 

949429 

108 

709349 

619 

290651 

53 

8 

669025 

411 

949364 

108 

709660 

619 

290340 

52 

9 

659271 

410 

949300 

108 

709971 

618 

290029 

51 

10 
11 

659517 

410 

949235 

108 
108 

710282 

518 

289718 

50 
49 

9.659763 

410 

9.949170 

9.710593 

518 

10.289407 

12 

660009 

409 

949105 

108 

710904 

518 

289096 

48 

13 

660255 

409 

949040 

108 

711215 

518 

288785 

47 

14 

660501 

409 

948975 

108 

711525 

.  517 

288475 

46 

15 

660746 

409 

948910 

108 

711836 

617 

288164 

45 

16 

660991 

408 

948845 

108 

712146 

517 

287864 

44 

17 

661236 

408 

948780 

109 

712456 

517 

287544 

43 

18 

661481 

408 

948715 

109 

712766 

516 

287234 

42 

19 

661726 

407 

948650 

109 

713076 

516 

286924 

41 

20 
21 

661970 
9.662214 

407 
407 

948584 

109 
109 

713386 

516 

286614 
10.286304 

40 
39 

9.948519 

9.713696 

516 

22 

662459 

407 

948454 

109 

714005 

516 

285995 

38 

23 

662703 

406 

948388 

109 

714314 

515 

285686 

37 

24 

662946 

406 

948323 

109 

714624 

616 

285376 

36 

25 

663190 

406 

948257 

109 

714933 

616 

28506? 

35 

26 

663433 

405 

948192 

109 

715242 

616 

284768 

34 

27 

663677 

405 

948126 

109 

716551 

614 

284449 

33 

28 

663920 

405 

948060 

109 

715860 

614 

284140 

32 

29 

664163 

405 

947995 

110 

716168 

614 

283832 

31 

30 
31 

664406 

404 

947929 

110 

716477 

614 

283523 

30 

29 

9.664648 

404 

9.947863 

9.716785 

614 

10.283216 

32 

664891 

404 

947797 

110 

717093 

513 

282907 

28 

33 

665133 

403 

947731 

110 

717401 

513 

282599 

27 

34. 

665375 

403 

947665 

110 

717709 

613 

282291 

26 

35 

665617 

403 

947600 

110 

718017 

513 

281983 

25 

36 

665859 

402 

947533 

no 

718326 

513 

281675 

24 

37 

666100 

402 

947467 

110 

718633 

512 

281367 

23 

38 

666342 

402 

947401 

110 

718940 

612 

281060 

22 

39 

666583 

402 

947335 

no 

719248 

512 

280762 

21 

40 
41 

666824 

401 

947269 

no 
no 

719566 

612 

280446 

20 
19 

9.667066 

401 

9.947203 

9.719862 

612 

10.280138 

42 

667306 

401 

947136 

111 

720169 

611 

279831 

18 

43 

667546 

401 

947070 

111 

720476 

611 

279624 

17 

44 

667786 

400 

947004 

111 

720783 

511 

279217 

16 

45 

668027 

400 

946937 

111 

721089 

611 

278911 

15 

46 

668267 

400 

946871 

111 

721396 

611 

278604 

14 

47 

668506 

399 

946804 

111 

721702 

610 

278298 

13 

48 

668746 

399 

946738 

111 

722009 

510 

277991 

12 

49 

668986 

399 

946671 

111 

722316 

510 

277686 

11 

60 

51 

669225 

399 

946604 
9.946538 

111 
111 

722621 

510 

277379 

10 
9 

9.669464 

398 

9.722927 

610 

10.277073 

52 

669703 

398 

946471 

111 

723232 

609 

276768 

8 

53 

669942 

398 

946404 

111 

723538 

609 

276462 

7 

54 

670181 

397 

946337 

111 

723844 

509 

276156 

6 

65 

670419 

397 

946270 

112 

724149 

609 

275851 

5 

66 

670658 

397 

946203 

112 

724454 

609 

275546 

4 

57 

670896 

897 

946136 

112 

724759 

508 

275241 

3 

58 

671134 

396 

946069 

112 

725065 

608 

274935 

2 

59 

671372 

396 

946002 

112 

726369 

508 

274631 

1 

60 

671609 

396 

945935 

112 

725674 

508 

274326 

0 

n 

Cosine 

Sine    1    i 

Cotang. 

Tang.  |M.| 

Degrees. 


46 

(28  Degrees.)  a 

TABLE  OF  LOGARITHMIC 

M. 

1    Sine 

D. 

I   Cosine  1  D. 

1   Taut?. 

1  n. 

Cotang.   1    1 

~~0 

9.671609 

396 

9.945935 

112 

9.725674 

508 

10.274323 

60 

1 

671847 

395 

945868 

112 

725979 

608 

274021 

69 

2 

672084 

395 

945800 

112 

726284 

607 

273716 

58 

3 

672321 

395 

945733 

112 

726688 

507 

273412 

67 

4 

672558 

395 

945666 

112 

726892 

507 

273108 

66 

6 

672795 

394 

945598 

112 

727197 

607 

272803 

55 

6 

673032 

394 

945531 

112 

727501 

507 

272499 

54 

7 

€73268 

394 

945464 

113 

727806 

506 

272195 

53 

8 

673505 

394 

945396 

113 

728109 

506 

271891 

52 

9 

673741 

393 

945328 

113 

728412 

606 

271588 

51 

10 

673977 

393 

945261 

113 

728716 

606 

271284 

50 

11 

9.674213 

393 

9.945193 

113 

9.729020 

606 

10.270980 

49 

12 

674448 

392 

946125 

113 

729323 

506 

270677 

48 

13 

674684 

•  392 

946058 

113 

729626 

605 

270374 

47 

14 

674919 

392 

944990 

113 

729929 

506 

270071 

46 

15 

675155 

392 

944922 

113 

730233 

605 

269767 

46 

16 

675390 

391 

944864 

113 

730535 

505 

269465 

44 

17 

675624 

391 

944786 

113 

730838 

504 

269162 

43 

18 

675859 

391 

944718 

113 

731141 

504 

268859 

42 

19 

676094 

391 

944660 

113 

731444 

604 

268566 

41 

20 
21 

676328 
9.676562 

390 

944582 

114 
114 

731746 

504 

268254 

40 
39 

390 

9.944514 

9.732048 

604 

10.267952 

22 

676796 

390 

944446 

114 

732351 

503 

267649 

38 

23 

677030 

390 

944377 

114 

732653 

603 

26731:7 

37 

24 

677264 

389 

944309 

114 

732955 

503 

267045 

36 

25 

677498 

389 

944241 

114 

733257 

503 

266743 

36 

26 

677731 

389 

944] 72 

114 

733568 

603 

266442 

34 

27 

677964 

388 

944104 

114 

733860 

502 

266140 

33 

28 

678197 

388 

944036 

114 

734162 

602 

266838 

32 

29 

678430 

388 

943967 

114 

734463 

502 

265537 

31 

30 
31 

678683 

388 

943899 

114 
114 

734764 

502 

265236 
10.264934 

30 
29 

9  678895 

387 

9.943830 

9.735066 

502 

32 

679128 

387 

943761 

114 

735367 

502 

264633 

28 

33 

679360 

387 

943693 

116 

73.5668 

601 

264332 

27 

34 

679592 

387 

943624 

115 

735969 

601 

264031 

26 

35 

679824 

386 

943656 

115 

736269 

501 

263731 

25 

36 

680056 

386 

943486 

116 

736670 

501 

263430 

24 

37 

680288 

386 

943417 

115 

736871 

501 

263129 

23 

38 

680519 

385 

943348 

115 

737171 

500 

262829 

22 

39 

680750 

385 

943279 

116 

737471 

500 

262529 

21 

40 
41 

680982 

386 

943210 

116 
115 

737771 
9.738071 

500 

262229 

20 
19 

9.681213 

385 

9.943141 

600 

10.261929 

42 

681443 

384 

943072 

115 

738371 

500 

261629 

18 

43 

681674 

384 

943003 

116 

738671 

499 

261329 

17 

44 

681905 

384 

942934 

116 

738971 

499 

261029 

16 

45 

682135 

384 

942864 

115 

739271 

499 

260729 

16 

46 

682365 

383 

942795 

116 

739670 

499 

260430 

14 

47 

682596 

383 

942726 

116 

739870 

499 

260130 

13 

48 

682825 

383 

942656 

116 

740169 

499 

259831 

12 

49 

683055 

383 

942587 

116 

740468 

498 

259532 

11 

50 
61 

683284 

382 

942517 

116 
116 

740767 

498 

259233 

10 
9 

9.683514 

382 

9.942448 

9.741066 

498 

10.258934 

S2 

683743 

382 

942378 

116 

741365 

498 

258635 

8 

63 

683972 

382 

942308 

116 

741664 

498 

268336 

7 

64 

684201 

381 

942239 

116 

741962 

49? 

268038 

6 

55 

684430 

381 

942169 

116 

742261 

497 

257739 

5 

56 

684658 

381 

942099 

116 

742559 

497 

257441 

4 

57 

684887 

380 

942029 

116 

742858 

497 

267142 

3 

58 

685115 

380 

941959 

il6 

743156 

497 

256844 

2 

59 

685343 

380 

941889 

117 

743454 

497 

256546 

1 

60 

685571 

380 

941819 

117 

743752 

496 

256248 

0 

Li 

Cosine   1 

Sine   1    1 

Colang. 

1    Tang.   1  M.  | 

61  Degrees. 


SINES  AND  TANGENTS 

.    ^29  E 

legrees 

•) 

47 

JL. 

Sine   1 

D. 

Cosine   1  D.  | 

Tang.   1 

D. 

Cotang.  1   1 

0 

9.685571 

380 

9.941819 

117 

9.743752 

496 

10.256248 

60 

1 

685799 

379 

941749 

117 

744050 

496 

255950 

59 

2 

686027 

379 

941679 

117 

744348 

496 

255652 

58 

3 

686254 

379 

941609 

117 

744645 

496 

255355 

57 

4 

686482 

379 

941539 

117 

744943 

496 

255057 

56 

5 

686709 

378 

941469 

117 

745240 

496 

254760 

55 

6 

686936 

378 

941398 

117 

745538 

495 

254462 

54 

7 

687163 

378 

941328 

117 

745835 

495 

254165 

53 

8 

687389 

878 

941258 

117 

746132 

495 

253868 

52 

9 

687616 

377 

941187 

117 

746429 

495 

253571 

51 

10 
11 

687843 
9.688069 

377 
377 

.941117 
9.941046 

117 
118 

746726 

495 

253274 
10.252977 

50 
49 

9.747023 

494 

12 

688295 

377 

940975 

118 

747319 

494 

252681 

48 

13 

688521 

376 

940905 

118 

747616 

494 

252384 

47 

14 

688747 

376 

940834 

118 

747913 

494 

252087 

46 

15 

688972 

376 

940763 

118 

748209 

494 

251791 

45 

16 

689198 

376 

940693 

118 

748505 

493 

251495 

44 

17 

689423 

375 

940622 

118 

748801 

493 

251199 

43 

18 

689648 

375 

940551 

118 

749097 

493 

250903 

42 

19 

689873 

375 

940480 

118 

749393 

493 

250607 

41 

20 
21 

690098 
9.690323 

375 
374 

940409 
9.940338 

118 
118 

749689 

493 

250311 

40 
39 

9.749985 

493 

10.250015 

32 

690548 

374 

940267 

118 

750281 

492 

249719 

38 

23 

690772 

374 

940196 

118 

750576 

492 

249424 

37 

24 

690996 

374 

940125 

119 

750872 

492 

249128 

36 

25 

691220 

373 

940054 

119 

751167 

492 

248833 

35 

26 

691444 

373 

939982 

119 

751462 

492 

248538 

34 

27 

691668 

373 

939911 

119 

751757 

492 

248243 

33 

28 

691892 

373 

939840 

119 

752052 

491 

247948 

32 

29 

692115 

372 

939768 

119 

752347 

491 

247653 

31 

30 
31 

692339 

372 

939697 
9.939625 

119 
119 

752642 

491 

247358 

,30 
29 

9.692562 

372 

9.752937 

491 

10.247063 

32 

692785 

371 

939554 

119 

753231 

491 

246769 

28 

33 

693008 

371 

939482 

119 

753526 

491 

246474 

27 

34 

693231 

371 

939410 

119 

758820 

490 

246180 

26 

35 

693453 

371 

939339 

119 

754115 

490 

245885 

25 

36 

693676 

370 

939267 

120 

754409 

490 

245591 

24 

37 

693898 

370 

939195 

120 

754703 

490 

245297 

23 

38 

694120 

370 

939123 

120 

754997 

490 

245003 

22 

39 

694342 

370 

939052 

120 

755291 

490 

244709 

21 

40 

41 

694564 
9.694786 

369 

938980 

120 
120 

755585 

489 

244415 

20 
19 

369 

9.938908 

9.755878 

489 

10.244122 

42 

695007 

309 

938636 

120 

756172 

489 

243828 

18 

43 

695229 

369 

938763 

120 

756465 

489 

243535 

17 

44 

695450 

368 

938691 

120 

756759 

489 

^43241 

16 

45 

695671 

368 

938619 

120 

757052 

489 

242948 

15 

46 

695892 

368 

938547 

120 

757345 

488 

242655 

14 

47 

696113 

368 

938475 

120 

757638 

488 

242362 

13 

48 

696334 

367 

938402 

121 

757931 

488 

242069 

12 

49 

696554 

367 

938330 

121 

758224 

488 

241776 

11 

50 
51 

696775 

367 

938258 

121 
121 

758517 

488 

241483 

10 
9 

9.696995 

367 

9.938185 

9.758810 

488 

10.241190 

52 

697215 

366 

938113 

121 

759102 

487 

240898 

8 

53 

697435 

366 

938040 

121 

759395 

487 

240605 

7 

54 

697654 

366 

937967 

121 

759687 

487 

240313 

6 

55 

697874 

366 

937895 

121 

759979 

487 

240021 

5 

56 

698094 

365 

937822 

121 

. 760272 

487 

239728 

4 

57 

698313 

365 

937749 

121 

760564 

487 

239436 

3 

58 

698532 

365 

937676 

121 

760856 

•  486 

239144 

2 

59 

698751 

365 

937604 

121 

761148 

486 

238852 

1 

60 

698970 

364 

937531 

121 

761439 

486 

238561 

0 

I  Cosine 

1  ■ 

1   Sine   1 

1  Cotang. 

1   Tang.   |M.| 

Degrees. 


48 

(30  Degr 

ees.)  A 

TABLE  OF  LOGARITHMIC 

T| 

Sine   1 

D.   1 

Cosine   1  D.  | 

'J'ansr.   | 

D.   1 

Cotang.   1   j 

0 

9.698970 

364 

9.937531  121 

9.761439 

486 

10.238561 

60 

1 

699189 

364 

937458  122 

761731 

486 

238269 

59 

2 

699407 

364 

937386  122 

762023 

486 

237977 

58 

3 

699626 

364 

937312 

122 

762314 

486 

237686 

57 

4 

699844 

363 

937238 

122 

762606 

485 

237394 

56 

5 

700062 

363 

937166 

122 

762897 

485 

237103 

55 

6 

700280 

363 

937092 

122 

763188 

485 

236812 

64 

7 

700498 

363 

937019 

122 

763479 

486 

236521 

53 

8 

700716 

363 

936946 

122 

763770 

485 

236230 

52 

9 

700933 

362 

936872 

122 

764061 

485 

236939 

51 

10 

11 

701151 
9.701368 

362 

936799 
9.936725 

122 
122 

764362 

484 

235648 

50 
49 

362 

9.764643 

484 

10.235367 

12 

701585 

362 

936662 

123 

764933 

484 

236067 

48 

13 

701802 

361 

936578 

123 

766224 

484 

234776 

47 

14 

702019 

361 

936506 

123 

765614 

484 

234486 

46 

15 

702236 

361 

936431 

123 

765805 

484 

234196 

45 

16 

702452 

361 

936357 

123 

766096 

484 

233905 

44 

17 

702669 

360 

936284 

123 

766385 

483 

233016 

43 

18 

702885 

360 

936210 

123 

766676 

483 

233326 

42 

19 

703101 

360 

936136 

123 

766965 

483 

233036 

41 

20 

21 

703317 

360 

936002 
9.935988 

123 
123 

767255 

483 

232746 

40 
39 

9.703533 

359 

9.767645 

483 

10.232456 

22 

703749 

359 

935914 

123 

767834 

483 

232166 

38 

23 

703964 

359 

935840 

123 

768124 

482 

231876 

37 

24 

704179 

359 

935766 

124 

768413 

482 

231587 

36 

25 

704395 

359 

935692 

124 

768703 

482 

231297 

35 

26 

704610 

358 

935618 

124 

768992 

482 

231008 

34 

27 

704825 

358 

935543 

124 

769281 

482 

230719 

33 

28 

705040 

358 

935469 

124 

769570 

482 

230430 

32 

29 

705254 

358 

935396 

124 

769860 

481 

230140 

31 

30 
31 

705469 
9.705683 

357 

935320 
9.935246 

•124 
124 

770148 

481 

229852 

30 
29 

357 

9.770437 

481 

10.229563 

32 

705898 

357 

935171 

124 

770726 

481 

229274 

28 

33 

706112 

357 

936097 

124 

771016 

481 

228985 

27 

34 

706326 

.356 

936022 

124 

771303 

481 

228697 

26 

35 

706539 

356 

934948 

124 

771592 

481 

228408 

25 

36 

706753 

356 

934873 

124 

771880 

480 

228120 

24 

37 

706967 

356 

934798 

125 

772168 

480 

227832 

23 

38 

707180 

355 

934723 

125 

772467 

480 

227543 

22 

39 

707393 

355 

934649 

126 

772745 

480 

227265 

21 

40 
41 

707606 
9.707819 

355 
355 

934574 
9.934499 

125 
125 

773033 

480 

226967 
10.226679 

20 
19 

9.773321 

480 

42 

708032 

354 

934424 

126 

773608 

479 

226392 

18 

43 

708245 

354 

934349 

125 

773896 

479 

226104 

17 

44 

708458 

354 

934274 

125 

774184 

479 

225816 

16 

45 

708670 

354 

934199 

125 

774471 

479 

226529 

15 

46 

708882 

353 

934123 

125 

774759 

479 

225241 

14 

47 

709094 

353 

934048 

126 

776046 

479 

224954 

13 

48 

709306 

353 

933973 

126 

775333 

479 

224667 

12 

49 

709518 

353 

933898 

126 

775621 

478 

2243S'9 

11 

50 
51 

709730 

353 

933822 
9.933747 

126 
126 

776908 

478 

224092 

10 
9 

9  709941 

362 

9.776195 

478 

10.223805 

52 

710153 

352 

933671 

126 

776482 

478 

223518 

8 

63 

710364 

362 

933596 

126 

776769 

478 

223231 

7 

54 

710575 

362 

933520 

126 

777055 

478 

222945 

6 

55 

710786 

351 

933446 

126 

777342 

478 

222668 

5 

56 

710997 

351 

933369 

126 

777628 

477 

222372 

4 

57 

711208 

351 

93329,1 

126 

777915 

477 

222086 

3 

58 

711419 

351 

933217 

126 

778201 

477 

221799 

2 

59 

711629 

360 

9331411 126 

778487 

477 

221512 

1 

60 

711839 

350 

933066' 12C 

778774 

477 

221226 

0| 

1   Cosine 

1 

1   SU.e   1 

1  Cotang. 

1    

1   Tiing.   1  M. 

1 

59  Degrees. 


SINES  AND  TANGENTS 

.   (31  L 

)  agrees 

0 

49 

M. 

Sine 

D. 

Cosine  1  D. 

Tansr. 

D. 

Cotang.   1   1 

"T 

9  711839 

350 

9.933066 

126 

9.778774 

477 

10.221226 

60 

1 

712050 

350 

932990 

127 

779060 

477 

220940 

59 

2 

712260 

350 

932914 

127 

779346 

176 

220654 

58 

3 

712469 

349 

932838 

127 

779632 

476 

220368 

57 

4 

712679 

849 

932762 

127 

779918 

476 

220082 

56 

5 

712889 

349 

932685 

127 

780203 

476 

219797 

55 

6 

713098 

349 

932609 

127 

780489 

476 

219511 

54 

7 

713308 

349 

932533 

127 

780775 

476 

219225 

53 

8 

713517 

348 

932457 

127 

781060 

476 

218940 

52 

9 

713726 

348 

932380 

127 

781346 

475 

218654 

51 

10 
11 

713935 

348 

932304 
9.932228 

127 
127 

781631 
9.781916 

475 

218369 

50 
49 

9.714144 

348 

475 

10.218084 

12 

714352 

347 

932151 

127 

782201 

475 

217799 

48 

13 

714561 

347 

932075 

128 

782486 

475 

217514 

47 

14 

714769 

347 

931998 

128 

782771 

475 

217229 

46 

15 

714978 

347 

931921 

128 

783056 

475 

216944 

45 

16 

715186 

347 

931845 

128 

783341 

475 

216659 

44 

17 

715394 

346 

931768 

128 

783626 

474 

216374 

43 

18 

715602 

346 

931691 

128 

783910 

474 

216090 

42 

19 

715809 

346 

931614 

128 

784195 

474 

215805 

41 

20 

21 

716017 
9.716224 

346 

931537 
9.931460 

128 

128 

784479 

474 

215521 

40 
39 

345 

9.784764 

474 

10.215236 

22 

716432 

345 

931383 

128 

785048 

474 

214952 

38 

23 

716639 

345 

931306 

128 

785332 

473 

214668 

37 

24 

716846 

345 

931229 

129 

785616 

473 

214384 

36 

25 

717053 

345 

931152 

129 

785900 

473 

214100 

35 

26 

717259 

344 

931075 

129 

786184 

473 

213816 

34 

27 

717466 

344 

930998 

129 

786468 

473 

213532 

33 

28 

717673 

844 

930921 

129 

786752 

473 

213248 

32 

29 

717879 

344 

930843 

129 

787036 

473 

212964 

31 

30 
31 

718085 

343 

930766 

129 
129 

787319 

472 

212681 

30 

29 

9.718291 

343 

9.930688 

9.787603 

472 

10.212397 

32 

718497 

343 

930611 

129 

787886 

472 

212114 

28 

33 

718703 

343 

930533 

129 

788170 

472 

211830 

27 

34 

718909 

343 

930456 

129 

788453 

472 

211547 

26 

35 

719114 

342 

930378 

129 

788736 

472 

211264 

25 

36 

719320 

342 

930300 

130 

789019 

472 

210981 

24 

37 

719525 

342 

930223 

130 

789302 

471 

210698 

23 

38 

719730 

342 

930145 

130 

789585 

471 

210415 

22 

39 

719935 

341 

930067 

130 

789868 

471 

210132 

21 

40 
41 

720140 
9.720345 

341 

929989 
9.929911 

130 
130 

790151 

471 

209849 

20 
19 

341 

9.790433 

471 

10.209567 

42 

720549 

341 

929833 

130 

790716 

471 

209284 

18 

43 

720754 

340 

929755 

130 

790999 

471 

209001 

17 

U 

720958 

340 

929677 

130 

791281 

471 

208719 

16 

45 

721162 

340 

929599 

130 

791563 

470 

208437 

15 

46 

721366 

340 

929521 

130 

791846 

470 

208154 

14 

47* 

721570 

340 

929442 

130 

792128 

470 

207872 

13 

48 

721774 

339 

929364 

131 

792410 

470 

207590 

12 

49 

721978 

339 

929286 

131 

792692 

470 

207308 

11 

50 

51 

722181 
9.722385 

339 
339 

929207 

131 
131 

792974 

470 

207026 

10 
9 

9.929129 

9.793256 

470 

10.206744 

52 

722588 

339 

929050 

131 

793538 

469 

206462 

8 

53 

722791 

338 

928972 

131 

793819 

469 

206181 

7 

54 

722994 

338 

928893 

131 

794101 

469 

205899 

6 

55 

723197 

338 

928815 

131 

794383 

469, 

205617 

6 

56 

723400 

338 

928736 

131 

794664 

469 

205336 

4 

57 

723603 

337 

928657 

131 

794945 

469 

205055 

3 

58 

723805 

337 

928578 

131 

795227 

469 

204773 

2 

59 

724007 

337 

928499 

131 

795508 

468 

204492 

1 

60 

724210 

337 

928420 

131 

795789 

468 

204211 

0 

Lj 

Cosine 

1   Sine   1 

Cotang. 

1 

Tang. 

m 

Be 


53  Degrees 

G 


60 


(32  Degrees.)     a  table  of  Logarithmic 


M. 

1   Sine 

1   D. 

1   Cosine   |  D. 

1   Tang. 

1  ■■■- 

1   Cotang.  1   1 

~ 

9.724210 

337 

9.928420 

132 

9.795789 

\     468 

10.2042111  60| 

1 

724412 

337 

928342 

132 

796070 

468 

203930 

59 

2 

724614 

336 

928263 

132 

796351 

468 

203649 

58 

3 

724816 

336 

928183 

132 

796632 

468 

203368 

57 

4 

725017 

336 

928104 

132 

796913 

468 

203087 

56 

5 

725219 

336 

928025 

132 

797194 

468 

202806 

55 

6 

725420 

335 

927946 

132 

797475 

468 

202525 

54 

7 

725622 

335 

927867 

132 

797755 

468 

202245 

53 

8 

725823 

335 

927787 

132 

798036 

467 

201964 

52 

9 

726024 

335 

927708 

132 

798316 

467 

201684 

51 

10 
11 

726225 

335 

927629 

132 
132 

798596 
9.798877 

467 
467 

201404 

50 
49 

9.726426 

334 

9.927549 

10.201123 

12 

726626 

334 

927470 

133 

799157 

467 

200843 

48 

13 

726827 

334 

927390 

133 

799437 

467 

200563 

47 

14 

727027 

334 

927310 

133 

799717 

467 

200283 

46 

15 

727228 

334 

927231 

133 

799997 

466 

200003 

45 

16 

727428 

333 

927151 

133 

800277 

466 

199723 

44 

17 

727628 

333 

927071 

133 

800557 

466 

199443 

43 

18 

727828 

333 

926991 

133 

800836 

466 

199164 

42 

19 

728027 

333 

926911 

133 

801116 

466 

198884 

41 

20 
21 

728227 

333 

926831 

133 
133 

801396 

466 

198604 

40 
39 

9.728427 

332 

9.926751 

9.801675 

466 

10.198325 

22 

728626 

332 

926671 

133 

801955 

466 

198045 

38 

23 

728825 

332 

926591 

133 

802234 

465 

197766 

37 

24 

729024 

332 

926511 

134 

802513 

465 

197487 

36 

25 

729223 

331 

926431 

134 

802792 

465 

197208 

35 

26 

729422 

331 

926351 

134 

803072 

465 

196928 

34 

27 

729621 

331 

926270 

134 

803351 

465 

196649 

33 

28 

729820 

331 

926190 

134 

803630 

465 

196370 

32 

29 

730018 

330 

926110 

134 

803908 

465 

196092 

31 

30 
31 

730216 
9.730415 

330 

926029 

134 
134 

804187 

465 

195813 

30 
29 

330 

9.925949 

9.804466 

464 

10.195534 

32 

730613 

330 

925868 

134 

804745 

464 

195255 

28 

33 

730811 

330 

925788 

134 

805023 

464 

194977 

27 

34 

731009 

329 

925707 

134 

805302 

464 

194698 

26 

35 

731206 

329 

925626 

134 

805580 

464 

194420 

25 

36 

731404 

329 

925545 

135 

805859 

464 

194141 

24 

37 

731602 

329 

925465 

135 

806137 

464 

193863 

23 

38 

731799 

329 

925384 

135 

806415 

463 

193585 

22 

39 

731996 

328 

925303 

135 

806693 

463 

193307 

21 

40 
41 

732193 

328 

925222 
9.925141 

135 
135 

806971 
9.807249 

463 

193029 

20 
19 

9.732390 

328 

463 

10.192751 

42 

732587 

328 

925060 

135 

807527 

463 

192473 

18 

43 

732784 

328 

924979 

135 

807805 

463 

192195 

17 

44 

732980 

327 

924897 

135 

808083 

463 

191917 

16 

45 

733177 

327 

924816 

135 

808361 

463 

191639 

15 

46 

733373 

327 

924735 

136 

808638 

462 

191362 

14 

47 

733569 

327 

924654 

136 

808916 

462 

191084 

13 

48 

733765 

327 

924572 

136 

809193 

462 

190807 

12 

49 

733961 

326 

924491 

136 

809471 

462 

190529 

11 

50 
51 

734157 

326 

924409 

136 
136 

809748 

462 

190252 

10 
9 

9.734353 

326 

9.924328 

9.810025 

462 

10.189975 

52 

734549 

326 

924246 

136 

810302 

462 

189698!  81 

53 

734744 

325 

924164 

136 

810580 

462 

189420J  7| 

64 

734939 

325 

924083 

136 

810857 

462 

189143 

6 

55 

735135 

325 

924001 

136 

811134 

461 

188866 

5 

66 

735330 

325 

923919 

136 

811410 

461 

188590 

4 

57 

735525 

325 

923837 

136 

811687 

461 

188313 

3 

58 

735719 

324 

923755 

137 

811964 

461 

1880.36 

2 

59 

735914 

324 

923673 

137 

812241 

461 

187759 

1 

60 

736109 

324 

923591 

137 

812517 

461 

187483 

0 

1 

"  Cosine 

1 

Sine   1    1 

Cotang.  j 

Tang.   1 

"W 

57  Degrees. 


■^ 


SINES  AND  TANGEKTS 

.  (33  Degrees 

0 

51 

M. 

Sine 

D. 

Cosine  |  D. 

Tang. 

D. 

Cotang.  1 

0 

9.736109 

324 

9.923591 

137 

9.812517 

461 

10.187482  60 

1 

736303 

324 

923509 

137 

812794 

461 

187206 

59 

2 

736498 

324 

923427 

137 

81.3070 

461 

186930 

68 

3 

736692 

323 

923.345 

137 

813.347 

460 

186653 

57 

4 

736886 

323 

923263 

137 

813623 

460 

186377 

66 

ft 

737080 

323 

923181 

137 

813899 

460 

186101 

55 

6 

737274 

323 

923098 

137 

814175 

460 

185825 

64 

7 

737467 

323 

923016 

137 

814452 

460 

185548 

63 

8 

737661 

322 

922933 

137 

814728 

460 

185272 

62 

9 

737855 

322 

922851 

137 

815004 

460 

184996 

61 

10 
11 

738048 

322 

922768 
9.922686 

138 

138 

815279 

460 

184721 

60 
49 

9.738241 

322 

9.815555 

459 

10.184445 

12 

738434 

322 

922603 

138 

81,5831 

459 

184169 

48 

13 

738627 

321 

922520 

138 

816107 

459 

183893 

47 

14 

738820 

321 

922438 

138 

816382 

459 

183618 

46 

15 

739013 

321 

922355 

138 

816658 

459 

183342 

46 

16 

739206 

321 

922272 

138 

816933 

459 

183067 

44 

17 

739398 

321 

922189 

138 

817209 

459 

182791 

43 

18 

739590 

320 

922106 

138 

817484 

459 

182516 

42 

19 

739783 

320 

922023 

138 

817759 

459 

182241 

41 

20 

21 

739975 

320 

921940 

138 

139 

818035 

458 

181965 

40 
39 

9.740167 

320 

9.921857 

9.818310 

458 

10.181690 

22 

740359 

320 

921774 

139 

818585 

458 

181415 

38 

23 

740550 

319 

921691 

139 

818860 

458 

181140 

37 

24 

740742 

319 

921607 

1.39 

819135 

458 

180865 

36 

25 

740934 

319 

921524 

139 

819410 

458 

180590 

36 

26 

741125 

319 

921441 

139 

819684 

458 

180316 

34 

27 

741316 

319 

921357 

139 

819959 

458 

180041 

33 

28 

741508 

318 

921274 

139 

820234 

458 

179766 

32 

29 

741699 

318 

921190 

139 

820508 

457 

179492 

31 

30 
31 

741889 
9.742080 

318 

921107 
9.921023 

139 
139 

820783 

457 

179217 

30 

29 

318 

9,821057 

457 

10.178943 

32 

742271 

318 

920939 

140 

821332 

457 

178668 

28 

33 

742462 

317 

920856 

140 

821606 

457 

178394 

27 

34 

742652 

317 

920772 

140 

821880 

457 

178120 

26 

35 

742842 

317 

920688 

140 

822154 

457 

177846 

25 

36 

743033 

317 

920604 

140 

822429 

457 

177571 

24 

37 

743223 

317 

920520 

140 

822703 

457 

177297 

23 

38 

743413 

316 

920436 

140 

822977 

456 

177023 

22 

39 

743002 

316 

920352 

140 

823250 

456 

176750 

21 

40 

743792 

316 

920268 

140 

823524 

456 

176476 

20 

41 

9.743982 

316 

9.920184 

140 

9.823798 

456 

10.176202 

19 

42 

744171 

316 

920099 

140 

824072 

456 

175928 

18 

43 

744361 

315 

920015 

140 

824345 

456 

175655 

17 

44 

744550 

315 

919931 

141 

824619 

456 

176381 

16 

45 

744739 

315 

919846 

141 

824893 

456 

175107 

15 

46 

744928 

315 

919762 

141 

825166 

456 

174834 

14 

47 

745117 

315 

919677 

141 

825439 

455 

174561 

13 

48 

745306 

314 

919593 

141 

825713 

455 

174287 

12 

49 

745494 

314 

919508 

141 

825986 

455 

174014 

11 

60 

745683 

814 

919424 

141 

826259 

455 

173741 

10 

51 

9.745871 

314 

9.919339 

141 

9.826532 

455 

10.173468 

9 

62 

74G059 

314 

919254 

141 

826805 

455 

173195 

8 

53 

746248 

313 

919169 

141 

827078 

455 

172922 

7 

54 

746436 

313 

919085 

141 

827351 

455 

172649 

6 

55 

746624 

313 

919000 

141 

827624 

455 

172370 

5 

56 

746812 

313 

918915 

142 

827897 

454 

172103 

4 

57 

746999 

313 

918830 

142 

828170 

454 

171830 

3 

58 

747187 

312 

918745 

142 

828442 

454 

171558 

2 

59 

747374 

312 

918659 

142 

828715 

454 

171285 

1 

60 

747562 

312 

9185741  142 

828987 

454 

171013 

0 

n 

Cosine 

Sine   1 

Colaiig. 

j    Tang.   1  M. 

56  Degrees. 


52 

(34  Degrees.)  a 

TABLE  OP  LOGARITHMIC 

M. 

1   Sine 

1   D. 

1   Cosine  |  D. 

1   Tang. 

1   D. 

1    Cotang.  1   1 

0 

9.747562 

312 

9.918574 

142 

9.828987 

454 

I0.17l\0l3 

60 

1 

747749 

312 

918489 

142 

829260 

454 

170740 

59 

2 

747936 

312 

918404 

142 

829532 

454 

170468 

58 

3 

748123 

311 

918318 

142 

829805 

454 

170195 

57 

4 

748310 

311 

9182.33 

142 

830077 

454 

169923 

56 

5 

748497 

311 

918147 

142 

830349 

453 

169651 

55 

6 

748683 

311 

918062 

142 

830621 

453 

•169379 

54 

7 

.748870 

311 

917976 

143 

830893 

453 

169107 

53 

8 

749056 

310 

917891 

143 

831 1&5 

453 

168835 

52 

9 

749243 

310 

'  917805 

143 

831437 

453 

168563 

51 

10 
11 

749429 
9.749615 

.   310 
310 

917719 

143 
143 

831709 

453 

168291 

50 
49 

9.917634 

9.831981 

453 

10.168019 

12 

749801 

310 

917548 

143 

832253 

453 

167747 

48 

13 

749987 

309 

917462 

143 

832525 

453 

167475 

47 

14 

750172 

309 

947376 

143 

832796 

453 

167204 

46 

15 

750358 

309 

-  9ri7290 

143 

833068 

452 

166932 

45 

16 

750543 

309 

917204 

143 

833339 

452 

166661 

44 

17 

750729 

.309 

9iril8 

144 

833611 

452 

166389 

43 

18 

750914 

308 

917032 

144 

833882 

452 

166118 

42 

19 

751099 

308 

916946 

144 

834154 

452 

165846 

41 

20 
21 

751284 
9.751469 

308 
308 

916859 
9.916773 

144 
144 

834425 

4.52 

165575 

40 
39 

9.834696 

452 

10.165304 

22 

751654 

308 

916687 

144 

834967 

452 

165033 

38 

23 

751839 

308 

916600 

144 

835238 

452 

164762 

37 

24 

752023 

307 

916514 

144 

835509 

452 

164491 

36 

25 

752208 

307 

916427 

144 

835780 

451 

164220 

35 

26 

752392 

307 

916341 

144 

838051 

451 

163949 

34 

27 

752576 

307 

916254 

144 

836322 

451 

163678 

33 

28 

752760 

307 

916167 

145 

836593 

451 

163407 

32 

29 

752944 

306 

916081 

145 

836864 

451 

163136 

31 

30 
31 

753128 

306 

915994 

145 
145 

837134 
9.837405 

451 
451 

162866 

30 
29 

9.753312 

306 

9.915907 

10.162595 

32 

753495 

306 

915820 

145 

837675 

451 

162325 

28 

33 

753679 

306 

915733 

145 

837946 

451 

162054 

27 

34 

753862 

305 

915646 

145 

838216 

451 

161784 

26 

35 

754046 

305 

915559 

145 

838487 

450 

161513 

25 

36 

754229 

305 

915472 

145 

838757 

450 

161243 

24 

37 

754412 

305 

915385 

145 

839027 

450 

160973 

23 

38 

754595 

305 

915297 

145 

839297 

450 

160703 

22 

39 

754778 

304 

915210 

145 

839568 

450 

160432 

21 

40 
41 

754960 

304 

915123 
9.915035 

146 
146 

839838 

450 

160162 

20 
19 

9.755143 

304 

,9.840108 

450 

10.159892 

42 

755346 

'?c304 

~9 14948 

146 

840378 

450 

159622 

18 

43 

755508 

'-304 

914860 

146 

840647 

450 

159353 

17 

44 

755690 

304 

914773 

146 

840917 

449 

159083 

16 

45 

755872 

303 

914685 

146 

841187 

449 

158813 

15 

46 

756054 

303 

914598 

146 

841457 

449 

1.58543 

14 

47 

756236 

303 

914510 

146 

841726 

449 

158274 

13 

48 

756418 

303 

914422 

146 

841996 

449 

158004 

12 

49 

756600 

303 

914334 

146 

842266 

449 

157734 

11 

50 
51 

756782 

302 

914246 
9.914158 

147 

147 

842535 
9.842805 

449 

157465 

10 
9 

9.756963 

302 

449 

10.157195 

52 

757144 

302 

914070 

147 

843074 

449 

156926 

8 

53 

757326 

302 

913982 

147 

843343 

449 

156657 

7 

54 

757507 

302 

913894 

147 

843612 

449 

156388 

6 

55 

757688 

301 

913806 

147 

843882 

448 

156118 

5 

56 

757869 

301 

913718 

147 

844151 

448 

155849 

4 

57 

758050 

301 

913630 

147 

844420 

448 

155580 

3 

58 

758230 

301 

913541 

147 

844689 

448 

15.5311 

2 

59 

758411 

301 

913453 

147 

844958 

448 

155042 

1 

60 

758591 

301 

91.3365 

147 

845227 

448 

154773 

0 

Cosine  1 

1 

Sine   1    1 

Colang. 

1 

tang.   M.| 

55  Degrees. 


SINES  AND  TANGENTS 

(35  Degrees. 

) 

63 

M. 

Sine   1 

D. 

Cosine   1  P.  . 

Tanc. 

D. 

Cotang.   1   1 

"Q- 

9.758591 

301 

9.913365 

147 

9.845227 

448 

10.154773 

60 

1 

758772 

300 

913276 

147 

845496 

448 

154504 

59 

2 

758952 

300 

913187 

148 

845764 

448 

154236 

58 

3 

759132 

300 

913099 

148 

846033 

448 

153967 

57 

4 

759312 

300 

913010 

148 

846302 

448 

153698 

56 

5 

759492 

300 

912922 

148 

846570 

447 

153430 

55 

6 

759672 

299 

912833 

148 

846839 

447 

153161 

54 

7 

759852 

299 

912744 

148 

847107 

447 

152893 

53 

8 

760031 

299 

912655 

148 

847376 

447 

152624 

52 

9 

760211 

299 

912566 

148 

847644 

447 

152356 

51 

10 
11 

760390 

299 

912477 
9.912388 

148 
148 

847913 
9.848181 

447 

152087 

50 
49 

9.760569 

298 

447 

10.151819 

12 

760748 

298 

912299 

149 

848449 

447 

151551 

48 

13 

760927 

298 

912210 

149 

848717 

447 

151283 

47 

14 

761106 

298 

912121 

149 

848986 

447 

151014 

46 

15 

761285 

298 

912031 

149 

849254 

447 

150746 

45 

16 

761464 

298 

911942 

149 

849522 

447 

150478 

44 

17 

761642 

297 

911853 

149 

849790 

446 

150210 

43 

18 

761821 

297 

911763 

149 

850058 

446 

149942 

42 

19 

761999 

297 

911674 

149 

850325 

446 

149675 

41 

20 
21 

762177 

297 

911584 
9.911495 

149 
149 

850593 

446 

149407 

40 
39 

9.762356 

297 

9.850861 

446 

10,149139 

22 

762534 

296 

911405 

149 

851129 

446 

148871 

38 

23 

762712 

296 

911315 

150 

851396 

446 

148604 

37 

24 

762889 

296 

911226 

150 

851664 

446 

148336 

36 

26 

763067 

296 

911136 

150 

851931 

446 

148069 

35 

26 

763245 

296 

911046 

150 

852199 

446 

147801 

34 

27 

763422 

296 

910956 

150 

852466 

446 

147534 

33 

28 

763600 

295 

910866 

150 

852733 

445 

147267 

32 

29 

763777 

295 

910776 

150 

853001 

445 

146999 

31 

30 
31 

763954 

295 

910686 

150 
150 

853268 

445 

146732 
10,146465 

30 
29 

9.764131 

295 

9.910596 

9.853535 

445 

32 

764308 

295 

910506 

150 

853802 

445 

146198 

28 

33 

764485 

294 

910415 

150 

854069 

445 

145931 

27 

34 

764662 

294 

910325 

151 

854336 

445 

145664 

26 

35 

764838 

294 

910235 

151 

854603 

445 

145397 

25 

36 

765015 

294 

910144 

151 

854870 

445 

145130 

24 

37 

765191 

294 

910054 

151 

855137 

445 

144863 

23 

38 

765367 

294 

909963 

151 

855404 

445 

144596 

22 

39 

765544 

293 

909873 

151 

855671 

444 

144329 

21 

40 
41 

785720 

293 

909782 
9.909691 

151 
151 

855938 

444 

144062 

20 
19 

9.765896 

293 

9.856204 

444 

10,143796 

42 

766072 

293 

909601 

151 

856471 

444 

143529 

18 

43 

766247 

293 

909510 

151 

856737 

444 

143263 

17 

44 

766423 

293 

909419 

151 

857004 

444 

142996 

16 

45 

766598 

292 

909328 

152 

857270 

444 

142730 

15 

46 

766774 

292 

909237 

152 

857537 

444 

142463 

14 

47 

766949 

292 

909146 

152 

857803 

444 

142197 

13 

48 

767124 

292 

909055 

152 

858069 

444 

141931 

12 

49 

767300 

292 

908964 

152 

858336 

444 

141664 

11 

50 
51 

767475 

291 

^08873 
9.908781 

152 
152 

858602 

443 

141398 

10 
~9 

9.767649 

291 

9,858868 

443 

10.141132 

52 

767824 

291 

908690 

152 

859134 

■  443 

140866 

8 

53 

767999 

291 

908599 

152 

859400 

443 

140600 

7 

54 

768173 

291 

908507 

152 

859666 

443 

140334 

6 

55 

768348 

290 

908416 

153 

859932 

443 

140068 

5 

56 

768522 

290 

908324 

153 

860198 

443 

139802 

4 

57 

768697 

290 

908233 

153 

860464 

443 

139536 

3 

68 

768871 

290 

908141 

153 

860730 

443 

139270 

2 

59 

769045 

290 

908049 

153 

860995 

443 

139005 

1 

60 

769219 

290 

907958' 153 

861261 

443 

138739 

0 

y  Cosine 

„ 

1   Sine   1 

['  Uotang. 

1   Tang.   1  M.  | 

Ee 


54  Degrees. 


64 

(36  Degrees.)  a 

TABLE  OF  LouARITHMIC 

M. 

Sine  •  1 

D. 

Cosine   |  I). 

Turn. 

D. 

Cotansr.   |    | 

"o" 

9.7692J9 

290 

9.907958 

153 

9.861261 

443 

10.1387.39 

60 

1 

769393 

289 

907866 

153 

861527 

443 

138473 

59 

2 

769566 

289 

90V VV4 

153 

861792 

442 

138208 

58 

3 

769740 

289 

907682 

153 

862058 

442 

137942 

57 

4 

769913 

289 

907590 

153 

862323 

442 

137677 

56 

5 

770087 

289 

907498 

153 

862589 

442 

137411 

55 

6 

770260 

288 

907406 

153 

862854 

442 

137146 

54 

7 

770433 

288 

907314 

154 

863119 

442 

136S81 

53 

8 

770606 

288 

907222 

154 

863385 

442 

136615 

52 

9 

770779 

288 

907129 

154 

863650 

442 

136350 

51 

10 
11 

770952 

288 

907037 

154 
154 

863915 

442 

136085 

50 
49 

9.771125 

288 

9  906945 

9.864180 

442 

10.135820 

12 

771298 

287 

906852 

154 

864445 

442 

135555 

48 

13 

771470 

287 

906760 

154 

864710 

442 

135290 

47 

14 

771643 

287 

906667 

154 

864975 

441 

135025 

46 

15 

771815 

287 

906575 

154 

865240 

441 

134760 

45 

16 

771987 

287 

906482 

154 

865505 

441 

134495 

44 

17 

772159 

287 

906389 

155 

865770 

441 

134230 

43 

18 

772331 

286 

906296 

155 

866035 

441 

133965 

42 

19 

772503 

286 

906204 

155 

866300 

441 

133700 

41 

20 
21 

772675 

286 

906111 

155 
155 

866561 

441 

133436 

40 
39 

9.772847 

286 

9.906018 

9.866829 

441 

10.133171 

22 

773018 

286 

905925 

155 

867094 

441 

132906 

38 

23 

77.*?  190 

286 

905832 

155 

867358 

441 

132642 

37 

24 

773361 

285 

905739 

155 

867623 

441 

132377 

36 

25 

773533 

285 

905645 

155 

867887 

441 

132113 

35 

26 

773704 

285 

905552 

155 

868152 

440 

131848 

34 

27 

773875 

285 

905459 

155 

868416 

440 

131584 

33 

28 

774046 

285 

905366 

156 

868680 

440 

131320 

32 

29 

774217 

285 

905272 

156 

868945 

440 

131055 

31 

30 
31 

774388 

284 

905179 

156 
156 

869209 

440 

130791 
10.130527 

30 

29 

9.774558 

284 

9.905085 

9.869473 

440 

32 

774729 

284 

904992 

156 

869737 

440 

130263 

28 

33 

774899 

284 

904898 

156 

870001 

440 

129999 

27 

34 

775070 

284 

904804 

156 

870265 

440 

129735 

26 

35 

775240 

284 

904711 

156 

870529 

440 

129471 

25 

36 

775410 

283 

904617 

156 

870793 

440 

129207 

24 

37 

775580 

283 

904523 

156 

871057 

440 

128943 

23 

38 

775750 

283 

904429 

157 

871321 

440 

128679 

22 

39 

775920 

283 

904335 

157 

871.585 

440 

128415 

21 

40 
41 

776090 

283 

904241 

157 
157 

871849 

439 

128151 

20 
19 

9.776259 

283 

9.904147 

9.872112 

439 

10.127888 

42 

776429 

282 

904053 

157 

872376 

439 

127624 

18 

43 

776598 

282 

903959 

157 

872640 

439 

127360 

17 

44 

776768 

282 

903864 

157 

872903 

439 

127097 

16 

45 

776937 

282 

903770 

157 

873167 

439 

126833 

15 

46 

777106 

282 

903676 

157 

873430 

439 

1265/0 

14 

47 

777275 

281 

903581 

157 

873694 

439 

126306 

13 

48 

777444 

281 

903487 

157 

873957 

439 

126043 

12 

49 

777613 

281 

903392 

158 

874220 

439 

125780 

11 

50 
51 

777781 

281 
281 

903298 

158 
158 

874484 

439 

125516 

10 
9 

9.777950 

9.903203 

9.874747 

439 

10.12.52.53 

52 

778119 

281 

903108 

158 

875010 

439 

124990 

8 

53 

778287 

280 

903014 

158 

875273 

438 

124727 

7 

54 

778455 

280 

902919 

158 

875536 

438 

124464 

6 

55 

778624 

280 

902824 

158 

•  875800 

438 

124200 

6 

56 

778792 

280 

902729 

158 

876063 

438 

12,3937 

4 

57 

778960 

280 

902634 

158 

876326 

438 

123674 

3 

58 

779128 

280 

902539 

159 

876589 

438 

123411 

2 

59 

779295 

279 

902444 

159 

876851 

438 

123149 

1 

eo_ 

779463 

279 

9023491  159 

877114 

438 

122888 

0 

Ctwine 

1   Sine   1 

Cotang. 

1 

1    Tang.  |M.  1 

53  Degrees. 


SINES  AND  TANGENTS.  (37  Degrees.) 

55 

M. 

Sine 

D.   1 

Cosine  1  D. 

Tans?. 

D. 

CotariE.  1    1 

"T 

9.779463 

279 

9.902349 

159 

9. 877114 

438 

10.122886 

60 

1 

779631 

279 

902253 

159 

877377 

438 

122623 

59 

2 

779798 

279 

902158 

159 

877640 

438 

122360 

58 

3 

779966 

279 

902063 

159 

877903 

438 

122097 

57 

4 

780133 

279 

901967 

159 

878165 

438 

121835 

56 

5 

780300 

278 

901872 

159 

878428 

438 

121572 

55 

6 

780467 

278 

901776 

159 

878691 

438 

121309 

54 

7 

780634 

278 

901681 

1.59 

878953 

437 

121047 

53 

8 

780801 

278 

901585 

159 

879216 

437 

120784 

52 

9 

780968 

278 

901490 

159 

879478 

437 

120522 

51 

10 

11 

781134 

278 

901394 

160 
160 

879741 
9.880003 

437 

120259 

50 
49 

9.781301 

277 

9.901298 

437 

10.119997 

12 

781468 

277 

901202 

160 

880265 

437 

119735 

48 

13 

781634 

277 

901106 

160 

880528 

437 

119472 

47 

14 

781800 

277 

901010 

160 

880790 

437 

119210 

46 

15 

781966 

277 

900914 

100 

881052 

437 

118948 

45 

16 

782132 

277 

900818 

16-) 

881314 

437 

118686 

44 

17 

782298 

276 

900722 

160 

881576 

437 

118424 

43 

18 

782464 

276 

900626 

160 

881839 

437 

118161 

42 

19 

782630 

276 

900529 

160 

882101 

437 

117899 

41 

20 
21 

782796 

276 

900433 
9.900337 

161 
161 

882363 

436 

117637 

40 
39 

9.782961 

276 

9.882625 

436 

10.117375 

22 

783127 

276 

900240 

161 

882887 

436 

117113 

38 

23 

783292 

275 

900144 

161 

883148 

436 

116852 

37 

24 

783458 

275 

900047 

161 

883410 

436 

116590 

36 

25 

783623 

275 

899951 

161 

883672 

436 

116328 

35 

26 

783788 

275 

899854 

161 

883934 

436 

116066 

34 

27 

783953 

275 

899757 

161 

884196 

436 

115804 

33 

28 

-  784118 

275 

899660 

161 

884457 

436 

115543 

32 

29 

784282 

274 

899564 

161 

884719 

436 

115281 

31 

30 
31 

784447 

274 

899467 
9.899370 

162 
162 

884980 

430 

11.5020 

30 
29 

9.784612 

274 

9.885242 

436 

10.114758 

32 

784776 

274 

899273 

162 

885503 

436 

114497 

28 

33 

784941 

274 

899176 

162 

885765 

436 

114235 

27 

34 

785105 

274 

899078 

162 

886026 

436 

113974 

26 

35 

785269 

273 

898981 

162 

886288 

436 

113712 

25 

36 

785433 

273 

898884 

162 

886549 

435 

113451 

24 

37 

785597 

273 

898787 

162 

886810 

435 

113190 

23 

38 

785761 

273 

898689 

162 

887072 

435 

112928 

22 

39 

785925 

273 

898592 

162 

887333 

435 

112667 

21 

40 
41 

786089 

273 

898494 
9.898397 

163 
163 

887594 

435 

112406 

20 
19 

9.786252 

272 

9.887855 

435 

10.112145 

42 

786416 

272 

898299 

163 

888116 

435 

111884 

18 

43 

786579 

272 

898202 

163 

888377 

435 

111623 

17 

44 

786742 

272 

898104 

163 

888639 

435 

.   111361 

16 

45 

786906 

272 

898006 

163 

888900 

435 

lUlOO 

15 

46 

787069 

272 

897908 

163 

889160 

435 

110840 

14 

47 

787232 

271 

897810 

163 

889421 

435 

110579 

13 

48 

787395 

271 

897712 

163 

889682 

435 

110318 

12 

49 

787557 

271 

897614 

163 

889943 

435 

110057 

11 

60 
51 

787720 

271 

897516 
9.897418 

163 
164 

890204 

434 

109796 

10 
9 

9.787883 

271 

9.890465 

434 

10.109535 

52 

788045 

271 

897320 

164 

890725 

434 

109275 

8 

53 

788208 

271 

897222 

164 

890986 

434 

109014 

7 

54 

788370 

270 

897123 

164 

891247 

434 

108753 

6 

55 

788532 

270 

897025 

164 

891507 

434 

108493 

5 

56 

788694 

270 

896926 

164 

891768 

434 

108232 

4 

57 

788856 

270 

896828 

164 

892028 

434 

107972 

3 

58 

789018 

270 

890729 

164 

892289 

434 

107711 

2 

59 

789180 

270 

896631 

164 

892549 

434 

107451 

1 

60 

789342 

269 

896532 

164 

892810 

434 

107190 

0 

~ 

Cosine 

Sine   1 

Colang. 

1   Tang.   1   1 

Degrees 


66 

(38  Degrees.)  a 

TABLE  OF  LOGABITHMTC 

M. 

1   Sine 

D. 

1   Cosine   |  D. 

Tang. 

D. 

CotanEj.  1 

0 

9.789342 

269 

9.896532 

164 

9.892810 

434 

10.107190,60 

1 

789504 

269 

896433 

165 

893070 

434 

106930 

59 

2 

789665 

269 

896335 

165 

893331 

434 

106669 

58 

3 

789827 

269 

896236 

165 

893591 

434 

106409 

57 

4 

789988 

269 

896137 

165 

893851 

434 

106149 

56 

5 

790149 

269 

896038 

165 

894111 

434 

105889 

55 

6 

790310 

268 

895939 

165 

894371 

434 

105629 

54 

7 

790471 

268 

895840 

165 

894632 

433 

105368 

53 

8 

790632 

268 

895741 

165 

894892 

433 

105108 

52 

9 

790793 

268 

895641 

165 

895152 

433 

104848 

51 

10 
11 

790954 

268 

895542 

165 
166 

895412 
9.895672 

433 

104588 

50 

49 

9.791115 

268 

9.895443 

433 

10,104328 

12 

791275 

267 

895343 

166 

895932 

433 

104068 

48 

13 

791436 

267 

895244 

166 

896192 

433 

103808 

47 

14 

791596 

267 

895145 

166 

896452 

433 

103548 

46 

15 

791757 

267 

895045 

166 

896712 

433 

103288 

45 

16 

791917 

267 

894945 

166 

896971 

433 

103029 

44 

17 

792077 

267 

894846 

166 

897231 

433 

102769 

43 

18 

792237 

266 

894746 

166 

897491 

433 

102509 

42 

19 

792397 

266 

894646 

166 

897751 

433 

102249 

41 

20 

21 

792557 

266 

894546 
9.894446 

166 
167 

898010 

433 

101990 

40 
39 

9.792716 

266 

9.898270 

433 

10.101730 

22 

792876 

266 

894346 

167 

898530 

433 

101470 

38 

23 

793035 

266 

894246 

167 

898789 

433 

101211 

37 

24 

793195 

265 

894146 

167 

899049 

432 

100951 

36 

25 

793354 

265 

894046 

167 

899308 

432 

100692 

35 

26 

793514 

265 

893946 

167 

899568 

432 

100432 

34 

27 

793673 

265 

893846 

167 

899827 

432 

100173 

33 

28 

793832 

205 

893745 

167 

900086 

432 

099914 

32 

29 

793991 

265 

893645 

167 

900346 

432 

099654 

31 

30 
31 

794150 

264 

893544 

167 
168 

900605 

432 

099395 

30 

29 

9.794308 

264 

9.893444 

9.900864 

432 

10.099136 

32 

794467 

264 

893343 

168 

901124 

432 

098S76 

28 

33 

794626 

264 

893243 

168 

901383 

432 

098617 

27 

34 

794784 

264 

893142 

168 

901642 

432 

098358 

26 

35 

794942 

264 

893041 

168 

901901 

432 

098099 

25 

36 

795101 

264 

892940 

168 

902160 

432 

097840 

24 

37 

795259 

263 

892839 

168 

902419 

432 

097581 

23 

38 

795417 

263 

892739 

168 

902679 

432 

097321 

22 

39 

795575 

263 

892638 

168 

902938 

432 

097062 

21 

40 
41 

795733 

263 

892536 

168 
169 

903197 

431 

096803 

20 
19 

9.795891 

263 

9.892435 

9.903455 

431 

10.096545 

42 

796049 

263 

892334 

169 

903714 

431 

096286 

18 

43 

796206 

263 

892233 

169 

903973 

431 

096027 

17 

44 

796364 

262 

892132 

169 

904232 

431 

095768 

16 

45 

796521 

262 

892030 

169 

904491 

431 

095509 

15 

46 

796679 

262 

891929 

169 

904750 

431 

095250 

14 

47 

796836 

262 

891827 

169 

905008 

431 

094992 

13 

48 

796993 

262 

891726 

169 

905267 

431 

094733 

12 

49 

797150 

261 

891634 

169 

905526 

431 

.  094474 

11 

50 
51 

797307 

261 

891523 
9.891421 

170 
170 

905784 
9.906043 

431 

094216 

10 
9 

9.797464 

261 

431 

10.093957 

52 

797621 

261 

891319 

170 

906302 

431 

093698 

8 

53 

797777 

261 

891217 

170 

906560 

431 

093440 

7 

54 

797934 

261 

891115 

170 

906819 

431 

093181 

6 

55 

798091 

261 

891013 

170 

907077 

431 

092923 

5 

56 

798247 

261 

890911 

170 

907336 

431 

092664 

4 

57 

798403 

260 

890809 

170 

907594 

431 

092406 

3 

58 

798560 

260 

890707 

170 

907852 

431 

092148 

2 

59 

798716 

260 

890605 

170 

908111 

430 

091889 

1 

60 

798872 

260 

890503 

170 

908369 

430 

091631 

0 

~ 

Cosine 

Sine   1    1 

Coiang.  1 

Tang.   |M.  1 

Si  Degrees. 


SINES  AND  TANGENTS.   (39  I 

)egrees 

0 

67 

M. 

i    Sine 

1   I>. 

1   Cosine   j  D. 

!   Tang. 

D. 

1   Cotang.   1   1 

0 

9.798872 

260 

9.890503 

170 

9.908369 

430 

10.091631 

60 

1 

799028 

260 

890400 

171 

908628 

430 

091372 

59 

2 

799184 

260 

890298 

171 

908886 

430 

091114 

58 

3 

799339 

259 

890195 

171 

909144 

430 

090856 

57 

4 

799495 

259 

890093 

171 

909402 

430 

090598 

56 

5 

799651 

259 

889990 

171 

909660 

430 

090340 

55 

6 

799806 

259 

889888 

171 

909918 

430 

090082 

54 

7 

799962 

259 

889785 

171 

910177 

430 

089823 

53 

8 

800117 

259 

889682 

171 

910435 

430 

089565 

52 

9 

800272 

258 

889579 

171 

910693 

430 

089307 

51 

10 
11 

800427 

258 

889477 
9.889374 

171 
172 

910951 

430 

089049 

50 

49 

9.800582 

258 

9.911209 

430 

10.088791 

12 

800737 

258 

889271 

172 

91 1467 

430 

088533 

48 

13 

800892 

258 

889168 

172 

911724 

430 

088276 

47 

14 

801047 

258 

889064 

172 

911982 

430 

088018 

46 

.15 

801201 

258 

888961 

172 

912240 

430 

087760 

45 

16 

801356 

257 

888858 

172 

A912498 

430 

087502 

44 

17 

801511 

257 

888755 

172 

912756 

430 

087244 

43 

18 

801665 

257 

888651 

172 

913014 

429 

086986 

42 

19 

801819 

257 

888548 

172 

913271 

429 

086729 

41 

20 

801973 

257 

888444 

173 

913529 

429 

086471 

40 

21 

9.802128 

257 

9.888341 

173 

9.913787 

429 

10.086213 

39 

22 

802282 

256 

888237 

173 

914044 

429 

085956 

38 

23 

802436 

256 

888134 

173 

9143,02 

429 

085698 

37 

24 

802589 

256 

888030 

173 

914560 

429 

085440 

36 

25 

802743 

256 

887926 

173 

'914817 

429 

085183 

35 

26 

802897 

256 

887822 

173 

915075 

429 

084925 

34 

27 

803050 

256 

887718 

173 

915332 

429 

084G68 

33 

28 

803204 

256 

887614 

173 

91,5590 

429 

084410 

32 

29 

803357 

255 

887510 

173 

915847 

423 

084153 

31 

30 

80.3511 

255 

887406 

174 

916104 

429 

083896 

30 

31 

9.803664 

255 

9.887302 

174 

9.916362 

429 

10.083638 

29 

32 

803817 

255 

887198 

174 

916619 

429 

083.381 

28 

33 

803970 

255 

887093 

174 

916877 

429 

083123 

27 

34 

804123 

2.55 

886989 

174 

917134 

429 

082866 

26 

35 

804276 

254 

886885 

174 

917391 

429 

082609 

26 

36 

804428 

254 

886780 

174 

917648 

429 

082352 

24 

27 

804581 

254 

886676 

174 

917905 

429 

082095 

23 

38 

804734 

254 

886571 

174 

918163 

428 

081837 

22 

39 

804886 

254 

886466 

174 

918420 

420 

08^.580 

21 

40 
41 

805039 

254 

886362 
9.88.6257 

175 
175 

918677 

428 

081323 
10.081066 

20 
19 

9.805191 

254 

9.918934 

■^"M 

42 

805343 

253 

886152 

175 

919191 

42«^ 

080809 

18 

43 

805495 

253 

886047 

175 

919448 

428 

080552 

17 

44 

805647 

253 

885942 

175 

919705 

428 

080295 

16 

45 

805799 

253 

885837 

175 

919962 

428 

080038 

15 

46 

805951 

253 

885732 

175 

920219 

428 

079781 

14 

47 

806103 

253 

885627 

175 

920476 

428 

'   079524 

13 

48 

800254 

253 

8855S2 

175 

92073S 

428 

079267 

12 

49 

806406 

252 

885416 

175 

920990 

428 

079010 

11 

50 

51 

806557 
9.806709 

252 

885311 
9.88.5205 

176 
176 

921247 

428 

078753 

10 
9 

252 

9.921503 

428 

JO. 078497 

52 

806860 

252 

885100 

176 

921760 

428 

078240 

8 

53 

807011 

252 

884994 

176 

922017 

428 

077983 

7 

54 

807163 

252 

884889 

176 

922274 

428 

077726 

6 

55 

807314 

252 

884783 

176 

922530 

428 

077470 

5 

56 

807465 

251 

884677 

176 

922787 

428 

077213 

4 

57 

807615 

251 

884572 

176 

923044 

428 

076956 

3 

58 

807766 

251 

884466 

176 

923300 

428 

076700 

2 

59 

807917 

251 

884360 

176 

923557 

427 

076443 

1 

60 

808067 

251 

884254 

177 

923813 

427 

076187 

0 

_J[ 

Cosine  1 

Sine   1 

Colang. 

1 

Tanj,'. 

T 

50  Degrees. 

H 


58 

(40  Degr 

ees.')  A 

TABLE  OF  LOGAEITHMIC 

M. 

1   Sine 

1   D. 

1  Cosine   ]  D. 

Tang. 

D. 

Cotanp.   1   1 

"T 

"9.808067 

251 

9.884254 

177 

9.923813 

427 

10.076187 

60 

1 

808218 

251 

884148 

177 

924070 

427 

07593U 

59 

2 

808368 

251 

884042 

177 

924327 

427 

075673 

58 

3 

808519 

250 

883936 

177 

924583 

427 

075417 

57 

4 

808669 

250 

883829 

177 

924840 

427 

075180 

56 

5 

808819 

250 

883723 

177 

925096 

427 

074904 

55 

6 

808969 

250 

883617 

177 

925352 

427 

074648 

54 

7 

809119 

250 

883510 

177 

925609 

427 

07439] 

53 

8 

809269 

250 

883404 

177 

925865 

427 

074135 

52 

9 

809419 

249 

883297 

178 

926122 

427 

073878 

51 

10 
11 

809569 
9.809718 

249 

883191 
9.883084 

178 

178 

926378 

427 

073622 

50 
49 

249 

9.926634 

427 

10.073366 

12 

809868 

249 

.  882977 

178 

925890 

427 

073110 

48 

13 

810017 

249 

882871 

178 

927147 

427 

072853 

47 

14 

810167 

249 

882764 

178 

927403 

427 

072597 

46 

15 

810316 

248 

882657 

178 

927659 

427 

072341 

45 

16 

810465 

248 

882550 

178 

927915 

427 

072085 

44 

17 

810614 

248 

882443 

178 

928171 

427 

071829 

43 

18 

810763 

248 

882336 

179 

928427 

427 

071573 

42 

19 

810912 

248 

882229 

179 

928683 

427 

071317 

41 

20 
21 

811061 

248 

882121 
9.882014 

179 
179 

928940 
9.929196 

427 

427 

071060 
10.070804 

40 
39 

9.811210 

248 

22 

811358 

247 

881907 

179 

929452 

427 

070548 

38 

23 

811507 

247 

881799 

179 

929708 

427 

070292 

37 

24 

811655 

247 

881692 

179 

929964 

426 

070036 

36 

25 

811804 

247 

881584 

179 

930220 

426 

069780 

35 

26 

811952 

247 

881477 

179 

930475 

426 

069525 

34 

27 

812100 

247 

881369 

179 

930731 

426 

069269 

33 

28 

812248 

247 

881261 

180 

930987 

426 

069013 

32 

29 

812396 

246 

8811.53 

180 

931243 

426 

068757 

31 

30 
31 

812544 

246 

881046 
9.880938 

180 
180 

931499 
9.931755 

426 

068501 
10.068245 

30 
29 

9.812692 

246 

426 

32 

« 12840 

240 

880830 

180 

932010 

426 

067990 

28 

33 

812988 

246 

880722 

180 

932266 

426 

067734 

27 

34 

813135 

246 

880613 

180 

932522 

426 

067478 

26 

35 

813283 

246 

880505 

180 

932778 

426 

067222 

25 

36 

813430 

245 

880397 

180 

933033 

426 

066967 

24 

37 

813578 

245 

880289 

181 

933289 

426 

066711 

23 

38 

813725 

245 

880180 

181 

933545 

426 

06.6455 

22 

39 

813872 

245 

880072 

181 

933800 

426 

066200 

21 

40 

41 

814019 
9.814166 

245 
f45 

879963 
9.8798,55 

181 
181 

934056 

426 

065944 

20 
19 

9.934311 

426 

10.065689 

42 

814313 

879746 

181 

934567 

426 

065433 

18 

43 

814460 

244 

879637 

181 

934823 

426 

065177 

17 

44 

814607 

244 

879529 

181 

935078 

426 

064922 

16 

45 

814753 

244 

879420 

181 

935333 

426 

064667 

15 

46 

814900 

244 

879311 

181 

935589 

426 

064411 

14 

47 

815046 

244 

879202 

182 

935844 

426 

064156 

13 

48 

815193 

244 

879093 

183 

936100 

426 

060900 

IS 

49 

815339 

244 

878984 

182 

936355 

426 

063645 

11 

50 
51 

815485 

243 

878875 
9.878766 

182 
182 

936610 

426 

063390 

10 
9 

9.815631 

243 

9.9.36866 

425 

10.063134 

52 

815778 

243 

878656 

182 

937121 

425 

062879 

8 

53 

815924 

243 

878547 

182 

937376 

425 

062624 

7 

54 

816069 

243 

878438 

182 

937632 

425 

062368 

6 

55 

816215 

243 

878328 

182 

937887 

425 

062113 

6 

56 

816361 

243 

878219 

183 

938142 

425 

061858 

4 

57 

816507 

242 

878109 

183 

938398 

425 

061602 

3 

58 

816652 

242 

877999 

183 

938653 

425 

061347 

2 

59 

816798 

242 

877890 

183 

938908 

425 

061092 

1 

60 

816943 

242 

877780!  183 

939163 

425 

060837 

0 

Cosine 

1   Sine   1 

Cotang, 

Tang.    M.  | 

49  Degrees. 


SINES  AND  TANGENTS 

.  (41  Degrees.) 

69 

M. 

Sine 

1   D. 

Cosine  |  D. 

Tann. 

D. 

Cotanfj.  1 

0 

9.816943 

24^ 

9.877780 

183 

9.939163 

425 

10.060837 

60 

1 

817088 

242 

877670 

183 

939418 

425 

060682 

59 

2 

817233 

242 

877560 

183 

939673 

425 

060327 

58 

3 

817379 

242 

877460 

183 

939928 

426 

060072 

67 

4 

817524 

241 

877340 

183 

940183 

425 

059817 

56 

6 

817668 

241 

877230 

184 

940438 

425 

059562 

65 

6 

817813 

241 

877120 

184 

940694 

425 

059306 

54 

7 

817958 

241 

877010 

184 

940949 

425 

059051 

53 

8 

818103 

241 

876899 

184 

941204 

425 

058796 

52 

9 

818247 

241 

876789 

184 

941468 

426 

058542 

51 

10 
11 

818392 

241 

876678 

184 
184 

941714 

426 

058286 

50 
49 

9.818636 

240 

9.876568 

9.941968 

425 

10.058032 

12 

818681 

240 

876457 

184 

942223 

425 

057777 

48 

13 

818826 

240 

876.347 

184 

942478 

425 

057522 

47 

14 

818969 

240 

876236 

185 

942733 

425 

057267 

46 

15 

819113 

240 

876125 

186 

942988 

426 

057012 

45 

16 

819267 

240 

876014 

186 

943243 

425 

056767 

44 

17 

819401 

240 

876904 

186 

943498 

425 

066502 

43 

18 

819545 

239 

876793 

185 

943752 

426 

056248 

42 

19 

819689 

239 

875682 

185 

944007 

425 

055993 

41 

20 
21 

819832 

239 

875571 
9.875469 

186 
185 

944262 

425 

055738 
10.055483 

40 
39 

9.819976 

239 

9.944517 

425 

22 

820120 

239 

875348 

186 

944771 

424 

055229 

38 

23 

820263 

239 

876237 

186 

946026 

424 

054974 

37 

24 

820406 

239 

875126 

186 

945281 

424 

054719 

36 

25 

820650 

238 

875014 

186 

945536 

424 

054465 

35 

26 

820693 

238 

.  874903 

186 

945790 

424 

064210 

34 

27 

820836 

238 

874791 

186 

946045 

424 

053956 

33 

28 

820979 

238 

874680 

186 

946299 

424 

063701 

32 

29 

821122 

238 

874568 

186 

946564 

424 

053446 

31 

30 
31 

821266 

238 

874456 
9.874344 

186 
186 

946808 

424 

053192 
10.052937 

30 
29 

9.821407 

238 

9.947063 

424 

32 

821650 

238 

874232 

187 

947318 

424 

062682 

28 

33 

821693 

237 

874121 

187 

947672 

424 

052428  27 

34 

821835 

237 

874009 

187 

947826 

424 

0.52174 

26 

35 

821977 

237 

873896 

187 

948081 

424 

051919 

25 

36 

822120 

237 

873784 

187 

948336 

424 

051664 

24 

37 

822262 

237 

873672 

187 

948590 

424 

051410 

23 

38 

822404 

237 

873560 

187 

948844 

424 

051156 

22 

39 

822546 

237 

873448 

187 

949099 

424 

050901 

21 

40 
41 

822688 

236 

873336 

187 
187 

949353 
9.949607 

424 

050647 

20 
19 

9.822830 

236 

9.873223 

424 

10.050393 

42 

822972 

236 

873110 

188 

949862 

424 

0,50138 

18 

43 

823114 

236 

872998 

188 

950116 

424 

049884 

17 

44 

823255 

236 

872885 

188 

950370 

424 

049630 

16 

45 

823397 

236 

872772 

188 

950626 

424 

049375 

15 

46 

823539 

236 

872659 

188 

950879 

424 

049121 

14 

47 

823680 

235 

872647 

188 

951133 

424 

048867 

13 

48 

823821 

235 

872434 

188 

951388 

424 

048612 

12 

49 

823963 

236 

872321 

188 

951642 

424 

048358 

11 

50 
51 

824104 
9.824245 

235 

872208 
9.872096 

188 
189 

961896 

424 

048104 

10 
9 

236 

9.962150 

424 

10.047850 

52 

824386 

235 

871981 

189 

952405 

424 

047695 

8 

53 

824527 

235 

871868 

189 

952659 

424 

047341 

7 

54 

824668 

234 

871755 

189 

962913 

424 

047087 

6 

56 

824808 

234 

871641 

189 

953167 

423 

046833 

6 

56 

824949 

234 

871628 

189 

953421 

423 

046579 

4 

57 

825090 

234 

871414 

189 

953675 

423 

046325 

3 

58 

825230 

234 

871301 

189 

953929 

423 

046071 

2 

69 

826371 

234 

871187 

189 

964183 

423 

046817 

1 

60 

825511 

234 

871073 

190 

954437 

423 

045563 

0 

1  Cosine 

" 

«ine   1 

Cotaiig. 

Tang.   1  M.  | 

48  Degrees. 


60 

(4i 

I  Degrees.)  a 

TABLE  OF  LOGARITHMIC 

M. 

Sine 

D. 

Cosine   1  D. 

Tanc 

D. 

Cotanp.   1   1 

0 

9.825511 

234 

9.871073 

190 

9.954437 

423 

10.046663 

60 

1 

825651 

233 

870960 

190 

954691 

423 

046309 

69 

2 

825791 

233 

870846 

]90 

954945 

423 

045055 

58 

3 

825931 

233 

870732 

]90 

955200 

423 

044800 

67 

4 

826071 

233 

870618 

190 

955454 

423 

044646 

66 

5 

826211 

233 

870504 

190 

955707 

423 

044293 

55 

6 

826351 

233 

870390 

190 

955961 

423 

044039 

54 

7 

826491 

233 

870276 

190 

956215 

423 

043785 

53 

8 

826631 

233 

870161 

190 

966469 

423 

043531 

52 

9 

826770 

232 

870047 

191 

966723 

423 

043277 

51 

10 
11 

826910 

232 

869933 
9.869818 

191 
191 

966977 
9.967231 

423 

043023 
10.042769 

50 
49 

9.827049 

232 

423 

12 

827189 

232 

869704 

191 

957485 

423 

042515 

48 

13 

827328 

232 

869589 

191 

957739 

423 

042261 

47 

14 

827467 

232 

869474 

191 

957993 

423 

042007 

46 

15 

827600 

232 

869360 

191 

958246 

423 

041764 

45 

16 

827745 

232 

869245 

191 

958600 

423 

041600 

44 

17 

827884 

231 

869130 

191 

9587.54 

423 

041246 

43 

18 

828023 

231 

869015 

192 

969008 

423 

040992 

42 

19 

828162 

231 

868900 

192 

959202 

423 

040738 

41 

20 
21 

828301 

231 

868785 

192 
192 

959516 
9.959769 

423 
423 

040484 
10.040231 

40 
39 

9.828439 

231 

9.868670 

22 

828578 

231 

868555 

192 

960023 

423 

039977 

38 

23 

828716 

231 

868440 

192 

960277 

423 

039723 

37 

24 

828855 

230 

868324 

192 

960531 

423 

039469 

36* 

25 

828993 

230 

868209 

192 

960784 

423 

039216 

35 

26 

829131 

230 

868093 

192 

961038 

423 

038962 

34 

27 

829269 

230 

867978 

193 

961291 

423 

038709 

33 

28 

829407 

230 

867862 

193 

961545 

423 

038455 

32 

29 

829545 

2.30 

867747 

193 

961799 

423 

038201 

31 

30 
31 

829683 

230 

867631 
9.867515 

193 
193 

962052 

423 

037948 
10.037694 

30 
29 

9.829821 

229 

9.962306 

423 

32 

829959 

229 

867399 

193 

962560 

423 

037440 

28 

33 

830097 

229 

867283 

193 

962813 

423 

037187 

27 

34 

830234 

229 

867167 

193 

963067 

423 

036933 

26 

35 

830372 

229 

867051 

193 

963320 

423 

036680 

25 

36 

830509 

229 

866935 

194 

963674 

423 

036426 

24 

37 

830646 

229 

866819 

194 

963827 

423 

036173 

23 

38 

830784 

229 

.  866703 

194 

964081 

423 

036919 

22 

39 

830921 

228 

866586 

194 

964335 

423 

036665 

21 

40 
41 

831058 

228 

866470 

194 
194 

964688 

422 

036412 
10.036158 

20 
19 

9.831195 

228 

9.866353 

9.964842 

422 

42 

831332 

228 

866237 

194 

965095 

422 

034905 

18 

43 

831469 

228 

866120 

194 

965349 

422 

034651 

17 

44 

831606 

228 

866004 

195 

965602 

422 

034398 

16 

45 

831742 

228 

865887 

195 

966856 

422 

034145 

15 

46 

831879 

228 

865770 

195 

966109 

422 

0.33891 

14 

47 

832015 

227 

865653 

195 

966362 

422 

033638 

13 

48 

832162 

227 

865636 

195 

966616 

422 

033384 

12 

49 

832288 

227 

86.5419 

196 

966869 

422 

033131 

11 

50 
51 

832426 

227 

865302 
9.865185 

195 
196 

967123 

422 

032877 

10 
9 

9.832561 

227 

9.967376 

422 

10.032624 

52 

832697 

227 

865068 

195 

967629 

422 

032371 

8 

53 

832833 

227 

864950 

195 

967883 

422 

032117 

7 

54 

832969 

226 

864833 

196 

968136 

422 

031864 

6 

55 

833105 

226 

864716 

196 

968389 

422 

031611 

5 

56 

833241 

226 

864598 

196 

968643 

422 

031357 

4 

67 

833377 

226 

864481 

196 

968896 

422 

031104 

3 

58 

833512 

226 

864363 

196 

969149 

422 

030861 

2 

59 

833648 

226 

864245 

196 

969403 

422 

030597 

1 

60 

833783 

226 

864127' 196 

969656 

422 

030344 

0 

1   Cosine 

1   Sine   1 

Cotang. 

P  Tang.   1  M. 

47  Degrees. 


^m 

"  -"'i^" 


SINES  AND  TANGENTS 

.   (43  Degrees.) 

61 

M. 

Sine 

D. 

Cosine   |  D. 

TanR. 

D. 

Cotang.  1   1 

0 

9.833783 

226 

9.864127 

196 

9.969656 

422 

10.030344 

*60 

1 

833919 

225 

864010 

196 

969909 

422 

030091 

59 

2 

834054 

225 

863892 

197 

970162 

422 

029838 

58 

3 

834189 

225 

863774 

197 

970416 

422 

029584 

57 

4 

834325 

225 

863656 

197 

970669 

422 

029331 

56 

5 

834460 

225 

863538 

197 

970922 

422 

029078 

55 

6 

834595 

225 

863419 

197 

971175 

422 

028825 

54 

7 

834730 

225 

863301 

197 

971429 

422 

028571 

53 

8 

834865 

225 

863183 

197 

971682 

422 

028318 

52 

9 

834999 

224 

863064 

197 

971935 

422 

028065 

51 

10 
11 

835134 

224 

862946 

198 
198 

972188 
9.972441 

422 
422 

027812 
10.027559 

50 
49 

9.835269 

224 

9.862827 

12 

835403 

224 

862709 

198 

972694 

422 

027306 

48 

13 

835538 

224 

862590 

198 

972948 

422 

027052 

47 

14 

835672 

224 

862471 

198 

973201 

422 

026799 

46 

15 

835807 

224 

862353 

198 

973454 

422 

026546 

45 

16 

835941 

224 

862234 

198 

973707 

422 

026293 

44 

17 

836075 

223 

862115 

198 

973960 

422 

026040 

43 

18 

836209 

223 

861996 

198 

974213 

422 

025787 

42 

19 

836343 

223 

861877 

198 

974466 

422 

025534 

41 

20 
21 

836477 

223 

861758 

199 
199 

974719 

422 

025281 
10.025027 

40 
39 

9.836611 

223 

9.861638 

9.974973 

422 

22 

836745 

223 

861519 

199 

975226 

422 

024774 

38 

23 

836878 

223 

861400 

199 

975479 

422 

024521 

37 

24 

837012 

222 

861280 

199 

975732 

422 

024268 

36 

25 

837146 

222 

861161 

199 

975985 

422 

024015 

35 

26 

837279 

222 

861041 

199 

976238 

422 

023762 

34 

27 

837412 

222 

860922 

199 

976491 

422 

023509 

33 

28 

837546 

222 

860802 

199 

976744 

422 

023256 

32 

29 

837679 

222 

860682 

200 

970997 

422 

023003 

31 

30 
31 

837812 

222 

860562 
9.860442 

200 
200 

977250 
9.977503 

422 

022750 

30 
29 

9.837945 

222 

422 

10.022497 

32 

838078 

221 

860322 

200 

977756 

422 

022244 

28 

33 

838211 

221 

860202 

200 

978009 

422 

021991 

27 

34 

838344 

221 

860082 

200 

978262 

422 

021738 

26 

35 

838477 

221 

859962 

200 

978515 

422 

021485 

25 

36 

838610 

221 

859842 

200 

978768 

422 

021232 

24 

37 

838742 

221 

859721 

201 

979021 

422 

020979 

23 

38 

838875 

221 

859601 

201 

979274 

422 

020726 

22 

39 

839007 

221 

859480 

201 

979527 

422 

020473 

21 

40 
41 

839140 

220 

859360 
9.859239 

201 
201 

979780 

422 

020220 

20 
19 

9.839272 

220 

9.980033 

422 

10.019967 

42 

839404 

220 

859119 

201 

980286 

422 

019714 

18 

43 

839536 

220 

858998 

201 

980538 

422 

019462 

17 

44 

839668 

220 

858877 

201 

980791 

421 

019209 

16 

45 

839800 

220 

858756 

202 

981044 

421 

018956 

15 

40 

839932 

220 

858635 

202 

981297 

421 

018703 

14 

47 

840064 

219 

858514 

202 

981550 

421 

018450 

13 

48 

840196 

219 

858393 

202 

981803 

421 

018197 

12 

49 

840328 

219 

858272 

202 

982056 

421 

017944 

11 

50 
51 

840459 

219 

858151 
9.858029 

202 
202 

982309 

421 

017691 
10.017438 

10 
9 

9.840591 

219 

9.982562 

421 

52 

840722 

219 

857908 

202 

982814 

421 

017186 

8 

53 

840854 

219 

857786 

202 

983067 

421 

016933 

7 

54 

840985 

219 

857665 

203 

983320 

421 

016680 

6 

55 

841116 

218 

857543 

203 

983573 

421 

016427 

5 

56 

841247 

218 

857422 

203 

983826 

421 

016174 

4 

57 

841378 

218 

857300 

203 

984079 

421 

015921 

3 

58 

841509 

218 

857178 

203 

984331 

421 

015669 

2 

59 

841640 

218 

857056 

203 

984584 

421 

015416 

1 

60 

841771 

218 

856934 

203 

984837 

421 

0151631  0 

1  Cosine 

1 

1   Sine   1 

Cotaiig. 

1    Tang.   1  M. 

46 

Degrees. 

Hh 


62 

(44  Degrees.)  a 

TABLE  OP  LOGARITHMIC 

nn" 

Sine   1 

D.   1 

Cosine   |  D.  | 

Tuns.        1  D.   | 

Cotansr.   |   f 

0 

9.841771 

218 

9.856934  203} 

9.984837 

421 

10.015163 

60 

1 

841902 

218 

856812  203| 

985090 

421 

014910 

59 

2 

842033 

218 

856690 

204 

985343 

421 

014657 

58 

3 

842163 

217 

856568 

204 

985596 

421 

014404 

57 

4 

842294 

217 

856446 

204 

985848 

421 

014152 

56 

5 

842424 

217 

856323 

204 

986101 

421 

013899 

55 

6 

842555 

217 

85620 1 

204 

986354 

421 

013646 

54 

7 

842685 

217 

856078 

204 

980607 

421 

013393 

53 

8 

842815 

217 

855956 

204 

986800 

421 

013140 

52 

9 

842946 

217 

855833 

204 

987112 

421 

012888 

51 

10 
11 

843076 
9.843206 

217 

85.5711 

205 
205 

987365 
9^.987618 

421 

012635 

50 
49 

216 

9.855588 

421 

10.012382 

12 

843336 

216 

855465 

205 

987871 

421 

012129 

48 

13 

843466 
843595 

216 

855342 

205 

988123 

421 

0'11877 

47 

14 

216 

855219 

205 

988376 

421 

0-11624 

46 

15 

843725 

216 

855096 

205 

988629 

421 

011371 

45 

16 

843855 

216 

854973 

205 

988882; 

421 

011118 

44 

17 

843984 

216 

854850 

205 

989134 

421 

010866 

43 

18 

844114 

215 

854727 

206 

989387 

421 

010613 

42 

19 

844243 

215 

854603 

206 

989640 

421 

010360 

41 

20 

21 

844372 

215 

854180 
9.854356 

206 
206 

989893 
9^.990145 

421 

010107 

40 
39 

9.844502 

2i5 

421 

10.009855 

22 

844631 

215 

854233 

206 

990398 

421 

009602 

38 

23 

844760 

215 

854109 

206 

990651 

421 

009349 

37 

24 

844889 

215 

853986 

206 

990903 

421 

009097 

36 

25 

845018 

215 

853862 

206 

991156 

421 

008844 

35 

26 

845147 

215 

853738 

206 

991409 

421 

008591 

34 

27 

845276- 

214 

853614 

207 

991662 

421 

008338 

33 

28 

845405 

214 

853490 

207 

991914 

421 

008086 

32 

29 

845533 

214 

853366 

207 

992167 

421 

007833 

31 

30 
31 

845662 
9.845790 

214 

853242 
9.853118 

207 
207 

992420 

421 

0075SO 

30 

29 

214 

9.992672 

421 

10.007328 

32 

845919 

214 

852994 

207 

992925 

421 

007075 

28 

33 

846047 

214 

852869 

207 

993178 

421 

006823 

27 

34 

846175 

214 

852745 

207 

993430 

421 

006570 

26 

35 

846304 

214 

852620 

207 

993683 

421 

006317 

25 

36 

846432 

213 

852496 

208 

993936 

421 

006064 

24 

37 

846560 

213 

852371 

208 

994189 

421 

005811 

23 

38 

846688 

213 

852247 

208 

994441 

421 

005559 

22 

39 

840816 

213 

852122 

208 

994694 

421 

005306 

21 

40 

846944 

213 

851997 

208 

994947 

421 

005053 

20 

41 

9.847071 

213 

9.851872 

208 

9.995199 

421 

10.004801 

19 

42 

847199 

213 

851747 

208 

995452 

421 

004548 

18 

43 

847327 

213 

851622 

208 

995705 

421 

004295 

17 

44 

847454 

212 

851497 

209 

995957 

421 

004043 

16 

45 

847582 

212 

851372 

209 

996210 

421 

003790 

15 

46 

847709 

212 

851246 

209 

996463 

421 

003537 

14 

47 

847836 

212 

851121 

209 

996715 

421 

003285 

13 

48 

847964 

212 

850996 

209 

996968 

421 

003032 

12 

49 

848091 

212 

850870 

209 

997221 

421 

002779 

11 

50 

848218 

212 

850745 

209 

997473 

421 

002527 

10 

51 

9.848345 

212 

9.850619 

209 

9.997726 

421 

10.002274 

9 

52 

848472 

211 

850493 

210 

997979 

421 

002021 

8 

53 

848599 

211 

850368 

210 

998231 

421 

001769 

7 

54 

848726 

211 

850242 

210 

998484 

421 

001516 

6 

55 

848852 

211 

850116 

210 

9987371  421 

001263 

5 

56 

848979 

211 

849990 

210 

998989  421 

001011 

4 

57 

849106 

211 

849864 

210 

999242!  421 

000758 

3 

58 

843232 

211 

849738 

210 

999495  421 

000505 

2 

59 

849359 

211 

849611 

210 

999748  421 

000253 

1 

Igo 

849485 

211 

849485 

210 

10.000000  421 

OOOOOOi  0 

Cosine 

I   aine   1 

1   Co;'aiig.   1 

1   Tang.    I  M. 

45  Degrees. 


Notices  of  Harper's  Libraries  of  Standard  Works. 


THE  FAMILY  LIBRARY. 

"  The  Family  Library— A  title  which,  from  the  valuable  and  entertaining  matter 
the  collection  contains,  as  well  as  from  the  careful  style  of  its  execution,  it  well  de- 
serves. No  family,  indeed,  in  which  there  are  children  to  be  brought  up,  ought  to 
be  without  this  Library,  as  it  furnishes  the  readiest  resources  for  that  education  which 
ought  to  accompany  or  succeed  that  of  the  boarding-school  or  the  academy,  and  is 
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Review. 

"  We  have  repeatedly  borne  testimony  to  the  utiUty  of  this  work.  It  is  one  of  the 
best  that  has  ever  been  issued  from  the  American  press,  and  should  be  in  the  library 
of  every  family  desirous  of  treasuring  up  useful  knowledge." — Boston  Statesman. 

*'  The  Family  Library  presents,  in  a  compendious  and  convenient  lorm,  well- 
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Charleston  Gazette. 


FAMILY  CLASSICAL  LIBRARY. 

"  The  Family  Classical  Library  is  another  of  those  cheap,  useful,  and  elegant  works 
which  we  lately  spoke  of  as  forming  an  era  in  our  publisning  history." — Spectator. 

"  This  work,  published  at  a  low  price,  is  beautifully  got  up.  Though  to  profess  to 
be  content  with  translations  of  the  Classics  has  been  denounced  as  '  the  thin  disguise 
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guages, who  would  yet  like  to  know  what  was  thought  and  said  by  the  sages  and 
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"  We  see  no  reason  why  this  work  should  not  find  its  way  into  the  boudoir  of  the 
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This  work  cannot  fail  to  be  acceptable  to  youth  of  both  sexes,  as  well  as  to  a  large 
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tion."— GerUleman's  Magazine,  Dec.  1829. 


BOY'S  AND  GIRL'S  LIBRARY. 

This  course  of  publications  win  more  especially  embrace  such  works  as  are 
adapted,  not  to  the  extremes  of  early  childhood  or  of  advanced  youth,  but  to  that 
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when  the  trifles  of  the  nursery  and  the  simple  lessons  of  the  school-room  have 
ceased  to  exercise  their  beneficial  influence,  but  before  the  taste  for  a  higher  order 
of  mental  pleasure  haus  estabhshed  a  fixed  ascendency  in  their  stead.  In  the  selec- 
tion of  works  intended  for  the  rising  generation  in  this  plastic  period  of  their  exist- 
ence, when  the  elements  of  future  character  are  receiving  their  moulding  impress, 
the  publishers  pledge  themselves  that  the  utmost  care  and  scrupulosity  shall  be  ex- 
ercised. They  are  fixed  in  their  determination  that  nothing  of  a  questionable  tend- 
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purpose  of  instructing  the  thoughts,  regulating  the  passions,  and  settling  the  princi- 
ples of  the  young.  Several  interesting  numbers  of  this  Library  are  now  before  the 
public. 

LIBRARY  OF  SELECT  NOVELS. 

Fictitious  composition  is  now  admitted  to  form  an  extensive  and  important  por- 
tion of  Uterature.  Well-wrought  novels  take  their  rank  by  the  side  of  real  rarratives, 
and  are  appealed  to  as  evidence  in  all  questions  concerning  man.  In  them  the  cus- 
toms of  countries,  the  transitions  and  shades  of  character,  and  even  the  very  pecu- 
liarities of  costume  and  dialect  are  curiously  preserved. 

This  "  Library  of  Select  Novels"  will  embrace  none  but  such  as  have  received  the 
impress  of  general  approbation,  or  have  been  written  by  authors  of  established  char- 
acter; and  the  publishers  hope  to  receive  such  encouragement  from  the  public 
patronage  as  will  enable  them  in  the  course  of  time  to  produce  a  series  of  works  of 
uniform  appearance,  and  including  most  oi  the  really  valuable  novels  and  romances 
that  have  been  or  shall  be  issued  from  the  modern  English  and  American  press. 

Sixteen  works,  by  eminent  authors,  have  already  been  published  in  this  Library, 
which  are  sold  separately  cr  in  complete  sets. — For  the  titles  see  the  Catalogue. 


Miscellaneous   Woi-Tts  Published  hy  J.  df  J.  Harper. 

THE   FAMILY  I -IBRARY.— Embracing  the  Following  Works 
in  18mo.     With  Plates,  &c. 

Nos.  1,  2,  3,  containing  Milnaan's  History  of  the  Jews. — 4,  5.  Lockhart's  Life  of  Napo- 
leon Bonaparte. — 6.  Soulhey's  Life  of  Nelson. — 7.  Williams's  Life  of  Alexander  the 
Great. — 8.  Natural  History  of  Insects. — 9,  Gait's  Life  of  Lord  Byron. — 10.  Bush's 
Life  of  Mohammed. — 1 1 .  Scott's  Letters  on  Demonology  and  Witchcraft. — 12, 13.  Gleig's 
History  of  the  Bible. — 14.  Discovery  and  Adventure  in  the  Polar  Seas,  <fc.  By  Profes- 
sor Leslie,  Professor  Jameson,  and  Hugh  Murray,  Esq. — 15.  Croly's  Life  of  George 
the  Fourth. — 16.  Discovery  and  Adventure  in  Africa.  By  Prof.  Jameson,  James 
Wilson,  Esq.,  and  Hugh  Murray,  Esq. — 17,  18,  19.  Cunningham's  Livi^s  of  Eminent 
Painters  rmd  Sculptors. — 20.  James's  History  of  Chivalry  and  the  Crusades. — 21,  22. 
Bell's  Life  of  Mary  Queen  of  Scots. — 23.  Russell's  Ancient  and  Modern  Egypt. — 24. 
Fletcher's  History  of  Poland. — 25.  Smith's  Festivals,  Games,  and  Amusements. — 26. 
Brewster's  Life  of  Sir  Isaac  Newton. — 27.  Russell's  History  of  Palestine,  or  the  Holy 
Land. — 28.  Memes'  Memoirs  of  the  Empress  Joscjjhine. — 29.  The  Court  and  Camp 
of  Bonaparte. — 30.  Lives  of  Early  Navigators. — 31.  A  Description  of  Pitcaim^s  Island, 
&c. — 32.  Turner's  Sacred  History  of  the  World. — 33,  34.  Mrs.  Jameson's  Memoirs 
of  Celebrated  Female  Sovereigns. — 35,  36.  Landers'  Africa. — 37.  Abercrombie  on  the 
Intellectual  Powers,  tfc— 38,  39,  40.  St.  John's  Lives  of  Celebrated  Travellers.— Ai,  42. 
Lord  Dover's  Life  of  Frederic  II.  King  of  Prussia.— i3,  44.  Sketches  from  Venetian 
History. — 45,  46.  Thatcher's  Indian  Biography. — 47,  48,  49.  History  of  India. — 50. 
Brewstcfs  Letters  on  Natural  Magic. — 51,52.  Taylor's  History  of  Ireland — 53.  Dis- 
coveries on  the  Northern  Coasts  of  America.-  -54.  Humboldt's  Travels. — 55,  56,  Euler's 
Letters  on  Natural  Philosophy. — 57.  Mudie's  Chiide  to  the  Observation  of  Nature. — 58, 
Abercrombie  on  the  Philosmhy  of  the  Moral  Feelings. — 59.  Dick  on  the  Improvement 
of  Society. — 60,  James's  History  of  Charlemagne. — 61.  Russell's  History  of  Nubia 
and  Abyssinia. — 62,  63.  Russell's  Life  of  Oliver  Cromwell. — 64,  65,  Cunningham's 
Lives  of  Eminent  Painters,  df-c,  vols.  4  &  5. 

CLASSICAL  SERIES.— Nos.  1,  2,  containing  Xenophon,  (Anabasis  and  Cyro- 
paedia.) — 3,  4.  Leland's  Demosthenes. — 5,  Rose's  Sallust, — 6,  7.  Caesar's  Comment- 
aries.—8,  9,  10.   Cicero's  Orations,  Offices,  &c. 

DRAMATIC  SERIES.— 1,  2,  3,  containing  Massinger's  Plays.— 4,  5,  Ford's 
Plays, 


THE   BOY'S  AND    GIRL'S  LIBRARY,— Embracing  the  Fol- 
lowing Works  in  18mo.     With  Engravings. 

No.  1,  being  Lives  of  the  Apostles,  &c. — 2,3.  Swiss  Family  Robinson. — 4.  Sunday 
Evenings,  1st  vol. — 5.  Son  of  a  Genius. — 6.  Uncle  Philip  on  Natural  History. — 7,  8. 
Indian  Traits. — 9,  10,  11,  Tales  from  American  History. — 12.  The  Young  Crusoe, — 
13.  Sunday  Evenings,  2d  vol. — 14.  Perils  of  the  Sea — 15.  Female  Biography. — 16, 
Caroline  Westerley, — 17,  Clergyman's  Orphan. — 18.  Ornaments  Discovered. — 19, 
Sunday  Evenings,  3d  vol. — 20,  Uncle  Philip  on  Christianity. — 21.  Uncle  Philip  on 
the  Trees  of  America. 


LIBRARY  OF  SELECT  NOVELS.— Embracing  the  Following 

Popular  Works.     12mo. 

Nos.  1,  2.  Cyril  Thornton.— 3,  4.  The  Dutchman's  Fireside,— 5,  6.  The  Young 
Duke.— 7,  8.  Caleb  Williams.— 9,  10.  The  Club-Book.— 11,  12.  De  Vere.— 13,  14, 
The  Smuggler.—15,  16.  Eugene  Aram.— 17,  18.  Evelina.— 19,20.  The  Spy,— 21,  22, 
Westward  Ho !— 23,  24.  Tales  of  Glauber-Spa.— 25, 26,  Henry  Masterton-— 27,  28. 
Mary  of  Burgundy.— 29,  30.  Richelieu,— 31,  32.   Darnley, 


THEOLOGICAL   LIBRARY.— Embracing  the   Following 

Works.— With  Plates. 

No.  1.  Life  of  Wiclif. — 2.  Consistency  of  Revelation. — 3,  4.  Life  of  Luther. — 5,  6. 
Life  of  Cranmer. 

OO"  -^ny  one  of  the  above  Works  may  he  obtained  separately. 


CLASSICAL    WORKS. 


XENOPHON.  The  Anabasis  :  Translated  by  Edward  Spel- 
MAN,  Esq.  The  Cyhop^dia  :  Translated  by  the  Hon.  Maurice  Ashly  Coo- 
per.    In  3  vols.  18mo.     With  a  Portrait. 


SALLUST.     Translated  by  William  Rose,  M.D.     With  Notes. 
18  mo.     Portrait. 


C^SAR.  With  Hirtius's  Continuation.  Translated  by  Wil- 
liam Duncan.     2  vols.  ISmo.     Portrait. 

CICERO.  The  Orations  Translated  by  Duncan,  the  Offices  by 
Cockman,  and  the  Cato  and  Laelius  by  Melmoth.  In  3  vols.  ISmo.  With  a 
Portrait. 


THE  WORKS  OF  HORACE,  translated  literally  into  English 
Prose.     By  C.  Smart,  A.M.     In  2  vols.  ISmo. 


THE    ORATIONS    OF   DEMOSTHENES.     Translated  by 
Thomas  Leland,  D.D.     In  2  vols.  ISmo.  Portrait. 


MEDICINE   AND    SURGERY. 

LEXICON   MEDICUM;    OR,   MEDICAL  DICTIONARY; 

Containing  an  Explanation  of  the  Terms  in  Anatomy,  Botany,  Chemistry,  Ma- 
teria Medica,  Midwifery,  Mineralogy,  Pharmacy^  Physiology,  Praictice  of  Physic, 
Suvgery,  and  the  Various  Branches  of  Natural  Philosophy  connected  with  Me- 
dicine. By  Robert  Hooper,  M.D.  With  Additions  from  American  Authors, 
by  Samuel  Akbrly,   M.D.      Two  \olumes  in  one,  8vo. 

A  DICTIONARY  OF  PRACTICAL  SURGERY :  compre- 
hending all  the  most  Interesting  Improvements,  from  the  earliest  Times  down 
to  the  Present  Period  ;  an  Account  of  the  Instruments  and  Remedies  employed 
in  Surgery  ;  the  Etymology  and  Signification  of  the  Principal  Terms  ;  and 
Numerous  References  to  Ancient  and  Modern  Works.  By  Samuel  Cooper, 
M.D.  With  numerous  Notes  and  Additions,  embracing  all  the  Principal  Im- 
provements and  Greater  Operations  introduced  and  performed  by  American 
Surgeons.  By  David  Meredith  Reese,  M.D.    In  2  vols.  8vo. 


THE  STUDY  OF  MEDICINE. 

New  and  Improved  Edition.     In  press. 


By  John  Mason  Good,  M.D. 


DIRECTIONS  FOR  INVIGORATING  AND  PROLONGING 

LIFE  ;  OR,  THE  INVALID'S  ORACLE.  Containing  Peptic  Precepts, 
pointing  out  agreeable  and  effectual  Methods  to  prevent  and  relieve  Indigestion, 
and  to  regulate  and  strengthen  the  Action  of  the  Stomach  and  Bowels.  By 
William  Kitchenee,  M.D.  Revised  and  Improved,  by  Thomas  S.  Barrett, 
M.D.     inmo. 


HISTORICAL    WORKS 

PUBLISHED   BY 

J.  &   J.  HARPER,  No.  82    CLIFF-STREET.   NEW-YORK. 


THE  HISTORY  OF  THE  DECLINE  AND  FALL  OF  THE 
ROMAN  EMPIRE.  By  Edward  Gibbon,  Esq.  Complete  in  4  vols.  8vo. 
With  Engravings. 

This  Stereotyped  Eaition  of  Gibbon's  Rome  is  well  printed  on  a  good  sized  type, 
and  contains  the  necessary  Maps,  and  is,  in  all  respects,  perfect.  These  facts  are 
stated,  because  most  of  the  London  editions  new  offered  for  sale  in  this  country  are 
without  the  necessary  Maps,  &c.,  and  are  printed  on  a  type  so  small  that  it  is  in- 
jurious to  the  eyes  to  read  them.  Yet,  with  all  these  disadvantages,  they  are  sold  at 
a  higher  price  than  this  American  Edition. 


THE  HISTORY  OF  MODERN  EUROPE ;  with  a  View  of 
the  Progress  of  Society,  from  the  Rise  of  the  Modern  Kingdoms  to  the  Peace 
of  Paris,  in  1763.  By  William  Russell,  LL.D.  :  and  a 'Continuation  of  the 
History  to  the  Present  Time,  by  William  Jones,  Esq.  With  Annotations 
by  an  American.     In  3  vols.  8vo.     With  Engravings. 


THE  HISTORY  OF  THE  DISCOVERY  AND  SETTLE- 
MENT OF  AMERICA.  By  William  Robertson,  D.D.  With  an  Account 
of  his  Life  and  Writings.  To  which  are  added  Questions  for  the  Examination 
of  Students.  By  John  Frost,  A.M.  Complete  in  1  vol.  8vo.  With  a  Por- 
trait and  Engravings. 


THE  HISTORY  OF  THE  REIGN  OF  THE  EMPEROR 
CHARLES  V.  With  a  View  of  the  Progress  of  Society  in  Europe,  from  the 
Subversion  of  the  Roman  Empire  to  the  Beginning  of  the  Sixteenth  Century. 
By  William  Robertson,  D.D.  To  which  are  added  Questions  for  the  Exam- 
ination of  Students.  By  John  Frost,  A.M.  Complete  in  1  vol.  8vo.  With 
Engravings. 


THE  HISTORY  OF  SCOTLAND  during  the  Reigns  of  Queen 
Mary  and  of  King  James  VI.  till  his  Accession  to  the  Crown  of  England. 
With  a  Review  of  the  Scottish  History  previous  to  that  Period  ;  and  an  Appen- 
dix containing  Original  Letters.     To  which  is  affixed 

AN  HISTORICAL  DISQUISITION  CONCERNING  THE 
KNOWLEDGE  THE  ANCIENTS  HAD  OF  INDIA ;  and  the  Progress 
of  Trade  with  that  Country  prior  to  the  Discovery  of  the  Passage  to  it  by  the 
Cape  of  Good  Hope.  With  an  Appendix  containing  Observations  on  the  Civil 
Policy,  the  Laws  and  Judicial  Proceedings,  the  Arts,  the  Sciences,  and  Reli- 
gious Institutions  of  the  Indians.  By  William  Robertson,  D.D.  Complete 
in  1  vol.  8vo.     With  Engravings. 

Robertson's  works  are  also  sold  in  sets 


THE  HISTORY  OF  THE  AMERICAN  THEATRE.     By 

William  Dunlap,  Vice  President  of  the  National  Academy  of  Design.     In  1 
vol.  8vo. 


ANNALS  OF  TRYON  COUNTY;  or,  the  Border  Warfare  of 
New-York,  during  the  Revolution.     By  Wm.  W.  Campbell.    8vo. 


Works  on  Theology,  6fc.  Published  by  J.  <^  J.  Harper. 


LUTHER  AND  THE  LUTHERAN  REFORMATION.     By 
John  Scott,  A.M.     In  2  vols.  18mo.     With  Portraits. 


THE  LIFE  OF  ARCHBISHOP  CRANMER.     By  Charles 
Webb  Le  Bas,  A.M.     In  2  vols.  18mo.     With  a  Portrait. 

THE  LIFE  OF  WICLIF.     By  Charles  Webb  Le  Bas,  A.M. 
18mo.     With  a  Portrait. 

THE  CONSISTENCY  OF   THE  WHOLE  SCHEME  OF 

REVELATION  with  Itself  and  with  Human  Reason.     By  Philip  Nicholas 
Shuttleworth,  D.D.     ISmo. 


THE  HISTORY  OF  THE  BIBLE.  By  Rev.  G.  R.  Gleig, 
A.M.     In  2  vols.     ISmo.     With  Maps. 

SERMONS  ON  SEVERAL  OCCASIONS.  By  Rev.  John 
Wesley,  A.M.  Containing  a  Number  of  Sermons  never  before  published  in 
this  Country.     In  3  vols.  Svo. 

PRESENT  STATE  OF  CHRISTIANITY,  and  of  the  Mis- 

sionary  Establishments  for  its  Propagation  in  all  Parts  of  the  World.     Edited 
by  Fredertc  Shoeerl.     12mo. 

KEY  TO  THE  REVELATION.  In  thirty-six  Lectures,  tak- 
ing the  whole  Book  in  course.     By  Ethan  Smith.     12mo. 


MATHEMATICS. 


AN  ELEMENTARY  TREATISE  ON  MECHANICS.  Trans- 
lated from  the  French  of  M.  Boucharlat.  With  Additions  and  Emendations, 
designed  to  adapt  it  to  the  Use  of  the  Cadets  of  the  U.  S.  Military  Academy. 
By  Professor  Edward  H.  Courtenay.     Svo. 


ELEMENTS  OF  DESCRIPTIVE  GEOMETRY,  with  their 

Application    to    Spherical    Trigonometry,  Spherical  Projections,   and  Warped 
\  Surfaces.     By  Professor  Charlks  Davies.     Plates.     Svo. 


A  TREATISE  ON  SHADES  AND  SHADOWS,  AND  LIN- 
EAR PERSPECTIVE.     By  Professor  Charlfs  Davies.     Svo. 

ELEMENTS  OF  SURVEYING.     With  the  necessary  Tables. 
By  Professor  Charles  Davies.     ISmo. 


GIBSON'S  SURVEYING.     New  and  Improved  Edition.     By 
J.  Ryan.     Svo. 

A  TABLE  OF  LOGARITHMS,  of  Logarithmic  Sines,  and  a 
Traverse  Table.     12mo. 


NATURAL    HISTORY. 


THE  BOOK  OF  NATURE.  By  John  Mason  Good,  M.D., 
F.R.S.  To  which  is  now  prefixed,  a  Sketch  of  the  Author's  Life.  Complete 
in  one  volume,  8vo. 

"  This  work  is  certainly  the  best  philosophical  digest  of  the  kind  which  we  have 
seen." — Monthly  Review. 

NATURAL  HISTORY  ;  or,  Uncle  Philip's  Conversations  with 
the  Children  about  Tools  and  Trades  among  the  Inferior  Animals.  18mo. 
With  numerous  Engrayings. 


THE  NATURAL  HISTORY  of  INSECTS,     With  numerous 

Engravings.     18mo. 

The  study  of  Natural  History  is  at  all  times,  and  to  almost  every  person,  eminently 
pleasing  and  instructive :  the  object  in  this  admirable  volume  has  been  to  render  it 
doubly  captivating  by  the  plain  and  simple  style  in  which  it  is  treated,  and  by  the 
numerous  engravings  with  which  the  text  is  illustrated.  There  is  no  branch  of  this 
dehghtful  science  more  pleasing  than  that  which  exhibits  the  wonderful  goodness  and 
wisdom  of  the  Creator,  as  they  are  displayed  in  the  endless  varieties  of  insect  life — 
their  forms,  habits,  capacities  and  works — and  which  investigates  the  nature  and 
peculiarities  of  these  diminutive  tribes  of  animated  existence. 


A  POPULAR  GUIDE  TO  THE   OBSERVATION  OF  NA- 
TURE.    By  Robert  Mudie,  Esq.     ISmo.     With  Engravings. 


AN  OUTLINE  OF  THE  NATURAL  HISTORY  OF 
EGYPT.  By  Rev.  Michael  Russell,  LL.D.  [No.  23  of  the  Family  Li- 
brary.]    18mo. 

AN  OUTLINE  of  THe'nATURAL  HISTORY  of  PALES- 
TINE.    By  Rev.  M.  Russell,  LL.D.     [No.  27  Fam.  Lib.]     18mo. 


AN  OUTLINE  OF  THE  NATURAL  HISTORY  OF  NU- 
BIA AND  ABYSSINIA.  By  Rev.  M.  Russkll,  LL.D.  18mo.  No.  61 
of  the  Family  Library.]     Engravings. 

DESCRIPTIVE  SKETCHES  OF  THE  NATURAL  HIS- 
TORY OF  THE  NORTH  AMERICAN  REGIONS.  By  James  Wilson, 
Esq.     18mo.     [No.  63  of  the  Family  Library.]     Engravings. 


ILLUSTRATIONS  OF  THE  CLIMATE,  GEOLOGY,  AND 
NATURAL  HISTORY  OF  THE  POLAR  SEAS  AND  REGIONS  ;  with 
an  Account  of  the  Whale-Fishery.  By  Professors  Leslie  and  Jameson.  With 
Engravings.     [No.  14  of  the  Family  Library.]     18mo. 

ILLUSTRATIONS  OF  THE  ZOOLOGY,  BOTANY,  CLI- 
MATE, GEOLOGY,  AND  MINERALOGY  OF  BRITISH  INDIA.  By 
James  Wilson,  Esq.  R.  K.  Greville,  liL.D.  and  Professor  Jameson.  18mo. 
[Nos.  47,  48,  &  49  of  the  Family  Library.]     Engravings. 


ILLUSTRATIONS  OF  THE  GEOLOGY,  MINERALOGY, 

AND  ZOOLOGY  OF  AFRICA.     By  Professor  Jameson  and  James  Wilson, 
Esq.     [No.  16  of  the  Family  Library.]     18mo. 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 
BERKELEY 

Return  to  desk  from  which  borrowed. 
This  book  is  DUE  on  the  last  date  stamped  below. 


SEP2  91^ 


LD  21-100m-7,'52(A2528sl6)476 


mm'-' 


^-"■A: 


IP 


